# algebra precalculus – Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$

## The Question :

108 people think this question is useful

$$x,y,z >0$$, prove
$$\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} \geqslant \frac{x+y+z}{13}$$

Note:
Often Stack Exchange asked to show some work before answering the question. This inequality was used as a proposal problem for National TST of an Asian country a few years back. However, upon receiving the official solution, the committee decided to drop this problem immediately. They don’t believe that any students can solve this problem in 3 hour time frame.

Update 1: In this forum, somebody said that BW is the only solution for this problem, which to the best of my knowledge is wrong. This problem is listed as “coffin problems” in my country. The official solution is very elementary and elegant.

Update 2: Although there are some solutions (or partial solution) based on numerical method, I am more interested in the approach with “pencil and papers.” I think the approach by Peter Scholze in here may help.

Update 3: Michael has tried to apply Peter Scholze’s method but not found the solution yet.

Update 4: Symbolic expanding with computer is employed and verify the inequality. However, detail solution that not involved computer has not been found. Whoever can solve this inequality using high school math knowledge will be considered as the “King of Inequality”.

• I have no idea about the official solution. I try this problem for the past 3 years but not yet success. Even with brute force, I still cannot solve it. This shows level of insanity this problem has.
• A simple observation is that the inequality is homogeneous, so it suffices to prove the case $x + y + z = 1$. Wolfram helped me solve the mess of an equation system that arises out of Lagrange multipliers, so I am convinced that the inequality is true now. (I have no idea how to solve the system myself, so this doesn’t really count as a solution.)
• I believe that $\sum_{\mathrm{cyc}}\frac{x^4}{ay^3+bz^3}\geq\frac{x+y+z}{a+b}$, and why not $\sum_{\mathrm{cyc}}\frac{x_n^4}{a_1x_1^3+\ldots a_{n-1}x_{n-1}^3}\geq\frac{x_1+\ldots+x_{n-1}+x_n}{a_1+\ldots+a_{n-1}+a_n}$ for $n\geq3$. Maybe the general case could be somehow helpful…
• Any further ideas for changes of the tags here must be first proposed to me.
• @YuriNegometyanov: If you have a solution then why don’t you just post it? – I find this request for a bounty a bit strange, and to be honest, I wonder that a moderator complied with it. (It isn’t just for the hat, or is it?)

28 people think this answer is useful

A big problem we get around $$(x,y,z)=(0.822,1.265,1.855)$$.

The Buffalo Way helps:

Let $$x=\min\{x,y,z\}$$, $$y=x+u$$,$$z=x+v$$ and $$x=t\sqrt{uv}$$.

Hence, $$\frac{13}{5}\prod\limits_{cyc}(8x^3+5y^3)\left(\sum\limits_{cyc}\frac{x^4}{8x^3+5y^3}-\frac{x+y+z}{13}\right)=$$

$$=156(u^2-uv+v^2)x^8+6(65u^3+189u^2v-176uv^2+65v^3)x^7+$$
$$+2(377u^4+1206u^3v+585u^2v^2-1349uv^3+377v^4)x^6+$$
$$+3(247u^5+999u^4v+1168u^3v^2-472u^2v^3-726uv^4+247)x^5+$$
$$+3(117u^6+696u^5v+1479u^4v^2+182u^3v^3-686u^2v^4-163uv^5+117v^6)x^4+$$
$$+(65u^7+768u^6v+2808u^5v^2+2079u^4v^3-1286u^3v^4-585u^2v^5+181uv^6+65v^7)x^3+$$$$+3uv(40u^6+296u^5v+472u^4v^2-225u^2v^4+55uv^5+25v^6)x^2+$$
$$+u^2v^2(120u^5+376u^4v+240u^3v^2-240u^2v^3-25uv^4+75v^5)x+$$
$$+5u^3v^3(8u^4+8u^3v-8uv^3+5v^4)\geq$$
$$\geq u^5v^5(156t^8+531t^7+2t^6-632t^5-152t^4+867t^3+834t^2+299t+40)\geq0$$

Done!

For example, we’ll prove that $$6(65u^3+189u^2v-176uv^2+65v^3)\geq531\sqrt{u^3v^3},$$ which gives a coefficient $$531$$ before $$t^7$$ in the polynomial $$156t^8+531t^7+2t^6-632t^5-152t^4+867t^3+834t^2+299t+40.$$

Indeed, let $$u=k^2v$$, where $$k>0$$.

Thus, we need to prove that:
$$130k^6+378k^4-177k^3-352k^2+130\geq0$$
and by AM-GM we obtain: $$130k^6+378k^4-177k^3-352k^2+130=$$
$$=130\left(k^3+\frac{10}{13}k-1\right)^2+\frac{k}{13}(2314k^3+1079k^2-5576k+2600)\geq$$
$$\geq\frac{k}{13}\left(8\cdot\frac{1157}{4}k^3+5\cdot\frac{1079}{5}k^2+21\cdot\frac{2600}{21}-5576k\right)\geq$$
$$\geq\frac{k^2}{13}\left(34\sqrt{\left(\frac{1157}{4}\right)^8\left(\frac{1079}{5}\right)^5\left(\frac{2600}{21}\right)^{21}}-5576\right)>0.$$
We’ll prove that $$2(377u^4+1206u^3v+585u^2v^2-1349uv^3+377v^4)\geq2u^2v^2,$$ for which it’s enough to prove that:
$$377t^4+1206t^3+584t^2-1349t+377\geq0$$ or
$$t^4+\frac{1206}{377}t^3+\frac{584}{377}t^2-\frac{1349}{377}t+1\geq0$$ or
$$\left(t^2+\frac{603}{377}t-\frac{28}{29}\right)^2+\frac{131015t^2-69589t+9633}{142129}\geq0,$$ which is true because
$$69589^2-4\cdot131015\cdot9633<0.$$

21 people think this answer is useful This is a question of the symmetric type, such as listed in:

With a constraint $\;x+y+z=1\;$ and $\;x,y,z > 0$ . Sort of a general method to transform such a constraint into the inside of a triangle in 2-D has been explained at length in:

Our function $f$ in this case is:
$$f(x,y,z) = \frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} – \frac{1}{13}$$
And the minimum of that function inside the abovementioned triangle must shown to be greater or equal to zero. Due to symmetry – why oh why can it not be proved with Group Theory – an absolute minimum of the function is expected
at $(x,y,z) = (1/3,1/3,1/3)$. Another proof without words is attempted
by plotting a contour map of the function, as depicted. Levels (nivo) of these isolines are defined (in Delphi Pascal) as:

nivo := min + sqr(g/grens)*(max-min); { sqr = square ; grens = 20 ; g = 0..grens }


The whiteness of the isolines is proportional to the (positive) function values; they are almost black near the minimum and almost white near the maximum values.
Maximum and minimum values of the function are observed to be:

 0.00000000000000E+0000 < f < 4.80709198767699E-0002


The little $\color{blue}{\mbox{blue}}$ spot in the middle is where $\,0 \le f(x,y,z) < 0.00002$ .

14 people think this answer is useful

Too long for a comment.

The Engel form of Cauchy-Schwarz is not the right way:

$$\frac{(x^2)^2}{8x^3+5y^3}+\frac{(y^2)^2}{8y^3+5z^3}+\frac{(z^2)^2}{8z^3+5x^3} \geq \frac{(x^2+y^2+z^2)^2}{13(x^3+y^3+z^3)}$$

So we should prove that
$$\frac{(x^2+y^2+z^2)^2}{13(x^3+y^3+z^3)}\geq\frac{x+y+z}{13}$$

which is equivalent to
$$\frac{(x^2+y^2+z^2)^2}{(x^3+y^3+z^3)}\geq x+y+z$$
but by Cauchy-Schwarz again we have $$x+y+z=\frac{(x^2)^2}{x^3} +\frac{(y^2)^2}{y^3}+\frac{(z^2)^2}{z^3} \geq \frac{(x^2+y^2+z^2)^2}{(x^3+y^3+z^3)}$$

and the inequalities are in the wrong way.

7 people think this answer is useful

This is more an extended comment to the answer of @MichaelRozenberg than an answer by its own.
I used a short Maxima to confirm the equation derived by @MichaelRozenberg.
I used Maxima because it is open source.

Here is the Maxima script (statements are terminated by $or by ;):  "I use string to comment this file"$

"the flag display2d  controls
the display of the output. You can unset it (display2d:false), that makes it easy to copy
the maxima output to math.stackexchange"$"to make it easier to input the problem data we define to function g and f:"$

g(r,s):=(8*r^3+5*s^3);

f(r,s):=r^4/g(r,s);

"
the initial problem has the form
L(x,y,t)>=R(x,y,z)
but we subtract R(x,y,z) from this equation and
we state the problem in the form
term0>=0
where term0 is L(x,y,z)-R(x,y,z)
this is term0:
"$term0:f(x,y)+f(y,z)+f(z,x)-(x+y+z)/13; " Now we multiply the term0 by a positive fraction of the (positive) common denominator and get term1 that satisfies term1>=0 ratsimp does some simplification like cancelling "$

term1:13/5*g(x,y)*g(y,z)*g(z,x)*term0,ratsimp;

"
now we assume x=0 and v>=0
,y=x+u and ,z=x+v do these substitutions
"$term2:term1,y=x+u,z=x+v; " ratsimp(.,x) does some simplification and displays the term as polynomial of x "$

term3:ratsimp(term2,x);

for p:0 thru hipow(term3,x) do print (coeff(term3,x,p)*x^p);

"the lowerbound polynomial is given by @Michael Rozenberg";

lowerbound:u^5*v^5*(156*t^8+531*t^7+2*t^6-632*t^5-152*t^4+867*t^3+834*t^2+299*t+40);

"we use the expanded version of the lowerbound polynomial";

lb:lowerbound,expand;

"we want to avoid squareroots and therefore substitute u bei q^2 and v by w^2.
The expression sqrt(u*v) (see thhe proof of Michael Rozenberg) then can be replaced by q*w";

"We want to avoid squareroots and therefore substitute u bei q^2 and v by w^2.
The expression sqrt(u*v) (see thhe proof of Michael Rozenberg) then can be replaced by q*w.
The following loop checks for each exponent k, that the coefficient of the original polynomial
in x (adjusted by sqrt(u*v)^k) is larger than the coeffiecient of the lowerbound polynomial.
This value is called wdiff in the following.
We already mentioned that we do not use the original variable u and v but first transform
to q and w as described above and therefor the adjustment is (q*w)^k  instead of sqrt(u*v)^k.
wdiff is a homogenous polynom of degree 20. We devide by wand replace q/w by s
and get the polynomial poly with vrailbe s. For these polynomials we calculate the number
of roots greater than 0. This can be done bei the nroot function that uses  'sturm's theorem'
Then we  calculate the value of poly at 2. If this value is greate 0 and there are
no zeros greater 0 then wdiff is greater or equal 0 for all nonnegative q and w and therefore
for all nonegative u and v. This was what we wanted to proof.
We see that all polynomails are positive at 2 and also for all except for k=8 there are no zeros
greater than 0. For k=8 we have a zero with even multiplicity.
";

for k:0 thru 8 do (
coff_x:coeff(term3,x,k),
coeff_t:coeff(lb,t,k),
wdiff:ev(coff_x*(q*w)^k-coeff_t,u=q^2,v=w^2),
poly:ratsubst(s,q/w,expand(wdiff/w^20)),
nr:nroots(poly,0,inf),
print("==="),
print("k=",k),
print("coeff(term3, x,",k,")=",coff_x),
print("coeff(lb, t,",k,")=",coeff_t),
print("wdiff=",wdiff),
print("polynomial:",poly),
print("factors=",factor(poly)),
print("number of roots >0:",nr),
print("poly(2)=",ev(poly,s=2))
);

"finally we proof that the lowerbbound polynomial has no positive root and that
it is greater than 0 for t=1. Therefor it is greater or equal than 0 for all admissible values";

poly:ratcoeff(lowerbound,u^5*v^5);

poly,t=1;

nroots(poly,0,inf);



I ran the scrip on the Xmaxima console and get the following output.
I use this console with this rather ugly kind of output because it
can be simply copied and pasted to math.stackecchange.
A prettier output can be found here at an online version of Maxima

(%i1) display2d:false;
(%o1) false
(%i2)
(%i3) "I use string to comment this file"
(%i4) "the flag display2d  controls

the display of the output. You can unset it (display2d:false), that makes it easy to copy

the maxima output to math.stackexchange"
(%i5) "to make it easier to input the problem data

we define to function g and f:"
(%i6) g(r,s):=8*r^3+5*s^3
(%o6) g(r,s):=8*r^3+5*s^3
(%i7) f(r,s):=r^4/g(r,s)
(%o7) f(r,s):=r^4/g(r,s)
(%i8) "

the initial problem has the form

L(x,y,t)>=R(x,y,z)

but we subtract R(x,y,z) from this equation and

we state the problem in the form

term0>=0

where term0 is L(x,y,z)-R(x,y,z)

this is term0:

"
(%i9) term0:f(x,y)+f(y,z)+f(z,x)+(-(x+y+z))/13
(%o9) z^4/(8*z^3+5*x^3)+y^4/(5*z^3+8*y^3)+((-z)-y-x)/13+x^4/(5*y^3+8*x^3)
(%i10) "

Now we multiply the term0 by a positive fraction of the (positive) common denominator

and get term1 that satisfies

term1>=0

ratsimp does some simplification like cancelling

"
(%i11) ev(term1:(13*g(x,y)*g(y,z)*g(z,x)*term0)/5,ratsimp)
(%o11) (25*y^3+40*x^3)*z^7+((-40*y^4)-40*x*y^3-64*x^3*y+40*x^4)*z^6
+(40*y^6+39*x^3*y^3-40*x^6)*z^4
+(40*y^7-64*x*y^6+39*x^3*y^4+39*x^4*y^3-40*x^6*y
+25*x^7)
*z^3+((-40*x^3*y^6)-64*x^6*y^3)*z+25*x^3*y^7
-40*x^4*y^6+40*x^6*y^4+40*x^7*y^3
(%i12) "

now we assume x=0 and v>=0

,y=x+u and ,z=x+v do these substitutions

"
(%i13) ev(term2:term1,y = x+u,z = x+v)
(%o13) (x+v)^3*(40*(x+u)^7-64*x*(x+u)^6+39*x^3*(x+u)^4+39*x^4*(x+u)^3+25*x^7
-40*x^6*(x+u))
+25*x^3*(x+u)^7+(x+v)*((-40*x^3*(x+u)^6)-64*x^6*(x+u)^3)
+(x+v)^4*(40*(x+u)^6+39*x^3*(x+u)^3-40*x^6)-40*x^4*(x+u)^6+40*x^6*(x+u)^4
+(x+v)^6*((-40*(x+u)^4)-40*x*(x+u)^3+40*x^4-64*x^3*(x+u))
+(x+v)^7*(25*(x+u)^3+40*x^3)+40*x^7*(x+u)^3
(%i14) "

ratsimp(.,x) does some simplification and displays the term as polynomial of x

"
(%i15) term3:ratsimp(term2,x)
(%o15) (156*v^2-156*u*v+156*u^2)*x^8+(390*v^3-1056*u*v^2+1134*u^2*v+390*u^3)
*x^7
+(754*v^4-2698*u*v^3+1170*u^2*v^2
+2412*u^3*v+754*u^4)
*x^6
+(741*v^5-2178*u*v^4-1476*u^2*v^3
+3504*u^3*v^2+2997*u^4*v+741*u^5)
*x^5
+(351*v^6-489*u*v^5-2058*u^2*v^4
+546*u^3*v^3+4437*u^4*v^2
+2088*u^5*v+351*u^6)
*x^4
+(65*v^7+181*u*v^6-585*u^2*v^5
-1286*u^3*v^4+2079*u^4*v^3
+2808*u^5*v^2+768*u^6*v+65*u^7)
*x^3
+(75*u*v^7+165*u^2*v^6-675*u^3*v^5
+1416*u^5*v^3+888*u^6*v^2
+120*u^7*v)
*x^2
+(75*u^2*v^7-25*u^3*v^6-240*u^4*v^5
+240*u^5*v^4+376*u^6*v^3
+120*u^7*v^2)
*x+25*u^3*v^7-40*u^4*v^6+40*u^6*v^4
+40*u^7*v^3
(%i16) for p from 0 thru hipow(term3,x) do print(coeff(term3,x,p)*x^p)
25*u^3*v^7-40*u^4*v^6+40*u^6*v^4+40*u^7*v^3
(75*u^2*v^7-25*u^3*v^6-240*u^4*v^5+240*u^5*v^4+376*u^6*v^3+120*u^7*v^2)*x
(75*u*v^7+165*u^2*v^6-675*u^3*v^5+1416*u^5*v^3+888*u^6*v^2+120*u^7*v)*x^2
(65*v^7+181*u*v^6-585*u^2*v^5-1286*u^3*v^4+2079*u^4*v^3+2808*u^5*v^2+768*u^6*v
+65*u^7)
*x^3

(351*v^6-489*u*v^5-2058*u^2*v^4+546*u^3*v^3+4437*u^4*v^2+2088*u^5*v+351*u^6)
*x^4

(741*v^5-2178*u*v^4-1476*u^2*v^3+3504*u^3*v^2+2997*u^4*v+741*u^5)*x^5
(754*v^4-2698*u*v^3+1170*u^2*v^2+2412*u^3*v+754*u^4)*x^6
(390*v^3-1056*u*v^2+1134*u^2*v+390*u^3)*x^7
(156*v^2-156*u*v+156*u^2)*x^8
(%o16) done
(%i17) "the lowerbound polynomial is given by @Michael Rozenberg"
(%o17) "the lowerbound polynomial is given by @Michael Rozenberg"
(%i18) lowerbound:u^5*v^5
*(156*t^8+531*t^7+2*t^6-632*t^5-152*t^4+867*t^3+834*t^2
+299*t+40)
(%o18) (156*t^8+531*t^7+2*t^6-632*t^5-152*t^4+867*t^3+834*t^2+299*t+40)*u^5*v
^5
(%i19) "we use the expanded version of the lowerbound polynomial"
(%o19) "we use the expanded version of the lowerbound polynomial"
(%i20) ev(lb:lowerbound,expand)
(%o20) 156*t^8*u^5*v^5+531*t^7*u^5*v^5+2*t^6*u^5*v^5-632*t^5*u^5*v^5
-152*t^4*u^5*v^5+867*t^3*u^5*v^5+834*t^2*u^5*v^5
+299*t*u^5*v^5+40*u^5*v^5
(%i21) "we want to avoid suareroots and therefore substitute u bei q^2 and v by w^2.

The expression sqrt(u*v) (see thhe proof of Michael Rozenberg) then can be replaced by q*w"
(%o21) "we want to avoid suareroots and therefore substitute u bei q^2 and v by w^2.

The expression sqrt(u*v) (see thhe proof of Michael Rozenberg) then can be replaced by q*w"
(%i22) "We want to avoid suareroots and therefore substitute u bei q^2 and v by w^2.

The expression sqrt(u*v) (see thhe proof of Michael Rozenberg) then can be replaced by q*w.

The following loop checks for each exponent k, that the coefficient of the original polynomial

in x (adjusted by sqrt(u*v)^k) is larger than the coeffiecient of the lowerbound polynomial.

This value is called wdiff in the following.

We already mentioned that we do not use the original variable u and v but first transform

to q and w as described above and therefor the adjustment is (q*w)^k  instead of sqrt(u*v)^k.

wdiff is a homogenous polynom of degree 20. We devide by wand replace q/w by s

and get the polynomial poly with vrailbe s. For these polynomials we calculate the number

of roots greater than 0. This can be done bei the nroot function that uses  'sturm's theorem'

Then we  calculate the value of poly at 2. If this value is greate 0 and there are

no zeros greater 0 then wdiff is greater or equal 0 for all nonnegative q and w and therefore

for all nonegative u and v. This was what we wanted to proof.

We see that all polynomails are positive at 2 and also for all except for k=8 there are no zeros

greater than 0. For k=8 we have a zero with even multiplicity.

"
(%o22) "We want to avoid suareroots and therefore substitute u bei q^2 and v by w^2.

The expression sqrt(u*v) (see thhe proof of Michael Rozenberg) then can be replaced by q*w.

The following loop checks for each exponent k, that the coefficient of the original polynomial

in x (adjusted by sqrt(u*v)^k) is larger than the coeffiecient of the lowerbound polynomial.

This value is called wdiff in the following.

We already mentioned that we do not use the original variable u and v but first transform

to q and w as described above and therefor the adjustment is (q*w)^k  instead of sqrt(u*v)^k.

wdiff is a homogenous polynom of degree 20. We devide by wand replace q/w by s

and get the polynomial poly with vrailbe s. For these polynomials we calculate the number

of roots greater than 0. This can be done bei the nroot function that uses  'sturm's theorem'

Then we  calculate the value of poly at 2. If this value is greate 0 and there are

no zeros greater 0 then wdiff is greater or equal 0 for all nonnegative q and w and therefore

for all nonegative u and v. This was what we wanted to proof.

We see that all polynomails are positive at 2 and also for all except for k=8 there are no zeros

greater than 0. For k=8 we have a zero with even multiplicity.

"
(%i23) for k from 0 thru 8 do
(coff_x:coeff(term3,x,k),coeff_t:coeff(lb,t,k),
wdiff:ev(coff_x*(q*w)^k-coeff_t,u = q^2,v = w^2),
poly:ratsubst(s,q/w,expand(wdiff/w^20)),nr:nroots(poly,0,inf),
print("==="),print("k=",k),print("coeff(term3, x,",k,")=",coff_x),
print("coeff(lb, t,",k,")=",coeff_t),print("wdiff=",wdiff),
print("polynomial:",poly),print("factors=",factor(poly)),
print("number of roots >0:",nr),print("poly(2)=",ev(poly,s = 2)))
===
k= 0
coeff(term3, x, 0 )= 25*u^3*v^7-40*u^4*v^6+40*u^6*v^4+40*u^7*v^3
coeff(lb, t, 0 )= 40*u^5*v^5
wdiff= 25*q^6*w^14-40*q^8*w^12-40*q^10*w^10+40*q^12*w^8+40*q^14*w^6
polynomial: 40*s^14+40*s^12-40*s^10-40*s^8+25*s^6
factors= 5*s^6*(8*s^8+8*s^6-8*s^4-8*s^2+5)
number of roots >0: 0
poly(2)= 769600
===
k= 1
coeff(term3, x, 1 )=
75*u^2*v^7-25*u^3*v^6-240*u^4*v^5+240*u^5*v^4+376*u^6*v^3
+120*u^7*v^2
coeff(lb, t, 1 )= 299*u^5*v^5
wdiff=
q*w
*(75*q^4*w^14-25*q^6*w^12-240*q^8*w^10+240*q^10*w^8+376*q^12*w^6
+120*q^14*w^4)
-299*q^10*w^10
polynomial: 120*s^15+376*s^13+240*s^11-299*s^10-240*s^9-25*s^7+75*s^5
factors= s^5*(120*s^10+376*s^8+240*s^6-299*s^5-240*s^4-25*s^2+75)
number of roots >0: 0
poly(2)= 7074016
===
k= 2
coeff(term3, x, 2 )=
75*u*v^7+165*u^2*v^6-675*u^3*v^5+1416*u^5*v^3+888*u^6*v^2
+120*u^7*v
coeff(lb, t, 2 )= 834*u^5*v^5
wdiff=
q^2*w^2
*(75*q^2*w^14+165*q^4*w^12-675*q^6*w^10+1416*q^10*w^6+888*q^12*w^4
+120*q^14*w^2)
-834*q^10*w^10
polynomial: 120*s^16+888*s^14+1416*s^12-834*s^10-675*s^8+165*s^6+75*s^4
factors= 3*s^4*(40*s^12+296*s^10+472*s^8-278*s^6-225*s^4+55*s^2+25)
number of roots >0: 0
poly(2)= 27198192
===
k= 3
coeff(term3, x, 3 )=
65*v^7+181*u*v^6-585*u^2*v^5-1286*u^3*v^4+2079*u^4*v^3
+2808*u^5*v^2+768*u^6*v+65*u^7
coeff(lb, t, 3 )= 867*u^5*v^5
wdiff=
q^3*w^3
*(65*w^14+181*q^2*w^12-585*q^4*w^10-1286*q^6*w^8+2079*q^8*w^6
+2808*q^10*w^4+768*q^12*w^2+65*q^14)
-867*q^10*w^10
polynomial:
65*s^17+768*s^15+2808*s^13+2079*s^11-867*s^10-1286*s^9-585*s^7
+181*s^5+65*s^3
factors=
s^3*(65*s^14+768*s^12+2808*s^10+2079*s^8-867*s^7-1286*s^6-585*s^4
+181*s^2+65)
number of roots >0: 0
poly(2)= 59331624
===
k= 4
coeff(term3, x, 4 )=
351*v^6-489*u*v^5-2058*u^2*v^4+546*u^3*v^3+4437*u^4*v^2
+2088*u^5*v+351*u^6
coeff(lb, t, 4 )= -152*u^5*v^5
wdiff=
q^4*w^4
*(351*w^12-489*q^2*w^10-2058*q^4*w^8+546*q^6*w^6+4437*q^8*w^4
+2088*q^10*w^2+351*q^12)
+152*q^10*w^10
polynomial: 351*s^16+2088*s^14+4437*s^12+698*s^10-2058*s^8-489*s^6+351*s^4
factors= s^4*(351*s^12+2088*s^10+4437*s^8+698*s^6-2058*s^4-489*s^2+351)
number of roots >0: 0
poly(2)= 75549104
===
k= 5
coeff(term3, x, 5 )=
741*v^5-2178*u*v^4-1476*u^2*v^3+3504*u^3*v^2+2997*u^4*v+741*u^5

coeff(lb, t, 5 )= -632*u^5*v^5
wdiff=
q^5*w^5
*(741*w^10-2178*q^2*w^8-1476*q^4*w^6+3504*q^6*w^4+2997*q^8*w^2
+741*q^10)
+632*q^10*w^10
polynomial: 741*s^15+2997*s^13+3504*s^11+632*s^10-1476*s^9-2178*s^7+741*s^5
factors= s^5*(741*s^10+2997*s^8+3504*s^6+632*s^5-1476*s^4-2178*s^2+741)
number of roots >0: 0
poly(2)= 55645088
===
k= 6
coeff(term3, x, 6 )= 754*v^4-2698*u*v^3+1170*u^2*v^2+2412*u^3*v+754*u^4
coeff(lb, t, 6 )= 2*u^5*v^5
wdiff=
q^6*w^6*(754*w^8-2698*q^2*w^6+1170*q^4*w^4+2412*q^6*w^2+754*q^8)
-2*q^10*w^10
polynomial: 754*s^14+2412*s^12+1168*s^10-2698*s^8+754*s^6
factors= 2*s^6*(377*s^8+1206*s^6+584*s^4-1349*s^2+377)
number of roots >0: 0
poly(2)= 22786688
===
k= 7
coeff(term3, x, 7 )= 390*v^3-1056*u*v^2+1134*u^2*v+390*u^3
coeff(lb, t, 7 )= 531*u^5*v^5
wdiff= q^7*w^7*(390*w^6-1056*q^2*w^4+1134*q^4*w^2+390*q^6)-531*q^10*w^10
polynomial: 390*s^13+1134*s^11-531*s^10-1056*s^9+390*s^7
factors= 3*s^7*(130*s^6+378*s^4-177*s^3-352*s^2+130)
number of roots >0: 0
poly(2)= 4482816
===
k= 8
coeff(term3, x, 8 )= 156*v^2-156*u*v+156*u^2
coeff(lb, t, 8 )= 156*u^5*v^5
wdiff= q^8*w^8*(156*w^4-156*q^2*w^2+156*q^4)-156*q^10*w^10
polynomial: 156*s^12-312*s^10+156*s^8
factors= 156*(s-1)^2*s^8*(s+1)^2
number of roots >0: 2
poly(2)= 359424
(%o23) done
(%i24) "finally we proof that the lowerbbound polynomial has no positive root and that

it is greater than 0 for t=1. Therefor it is greater or equal than 0 for all admissible values"
(%o24) "finally we proof that the lowerbbound polynomial has no positive root and that

it is greater than 0 for t=1. Therefor it is greater or equal than 0 for all admissible values"
(%i25) poly:ratcoef(lowerbound,u^5*v^5)
(%o25) 156*t^8+531*t^7+2*t^6-632*t^5-152*t^4+867*t^3+834*t^2+299*t+40
(%i26) ev(poly,t = 1)
(%o26) 1945
(%i27) nroots(poly,0,inf)
(%o27) 0
(%i28)


Here we list the coefficient functions so we can compare them to @MichaelRozenbergs function to see they are the same.

$$\begin{array}{r} \tag{1} \left(25\,u^3\,v^7-40\,u^4\,v^6+40\,u^6\,v^4+40\,u^7\,v^3\right)\,x^0 \\ \left(75\,u^2\,v^7-25\,u^3\,v^6-240\,u^4\,v^5+240\,u^5\,v^4+376\,u^ 6\,v^3+120\,u^7\,v^2\right)\,x^1 \\ \left(75\,u\,v^7+165\,u^2\,v^6-675\,u^3\,v^5+1416\,u^5\,v^3+888\,u^ 6\,v^2+120\,u^7\,v\right)\,x^2 \\ \left(65\,v^7+181\,u\,v^6-585\,u^2\,v^5-1286\,u^3\,v^4+2079\,u^4\,v ^3+2808\,u^5\,v^2+768\,u^6\,v+65\,u^7\right)\,x^3 \\ \left(351\,v^6-489\,u\,v^5-2058\,u^2\,v^4+546\,u^3\,v^3+4437\,u^4\, v^2+2088\,u^5\,v+351\,u^6\right)\,x^4 \\ \left(741\,v^5-2178\,u\,v^4-1476\,u^2\,v^3+3504\,u^3\,v^2+2997\,u^4 \,v+741\,u^5\right)\,x^5 \\ \left(754\,v^4-2698\,u\,v^3+1170\,u^2\,v^2+2412\,u^3\,v+754\,u^4 \right)\,x^6 \\ \left(390\,v^3-1056\,u\,v^2+1134\,u^2\,v+390\,u^3\right)\,x^7 \\ \left(156\,v^2-156\,u\,v+156\,u^2\right)\,x^8 \end{array}$$

To proof that this function is larger than
$$\left(156\,t^8+531\,t^7+2\,t^6-632\,t^5-152\,t^4+867\,t^3+834\,t^2+ 299\,t+40\right)\,u^5\,v^5 \tag{2}$$
Rozenbergs’s lower bound when we substitute $x$ by $t\sqrt(uv)$ we show that each coefficient of the polynomial $(1)$ is larger than the corresponding coefficient of the lower bound polynomial $(2)$.
Then we show that the polynomial $(2)$ is larger than $0$ for all nonnegative $u$, $v$ and $t$. Details can be found in the Maxima script.

Instead of the Maxima nroots function, which is based on Sturm sequences, one could solve the equations by some numeric functions to see if there are zeros greater than zeros, e.g. calculating the roots of poly for k=7 gives the following:

(%i29) allroots(390*s^13+1134*s^11-531*s^10-1056*s^9+390*s^7 ,s);
(%o29) [s = 0.0,s = 0.0,s = 0.0,s = 0.0,s = 0.0,s = 0.0,s = 0.0,
s = 0.007444635413686057*%i+0.7516683014652126,
s = 0.7516683014652126-0.007444635413686057*%i,
s = 0.3202741285237583*%i-0.6047586795035632,
s = (-0.3202741285237583*%i)-0.6047586795035632,
s = 1.93839678615644*%i-0.1469096219616494,
s = (-1.93839678615644*%i)-0.1469096219616494]


So we can also conclude are no real roots greater than 0. But this method is not really acceptable if one does not analyze the impact of the rounding errors. And this can be very complicated. The nroots function works with integers (for integer polynomials) and so there are no rounding errors.

5 people think this answer is useful

I write a start for a full answer (this is an idea that @Starfall first proposed in comment). If someone wants to use it to end the proof, she/he is welcome!

Let
$$f(x,y,z):=\frac{x^4}{ax^3+by^3}+\frac{y^4}{ay^3+bz^3}+\frac{z^4}{az^3+bx^3}.$$
Since $f$ is homogeneous of degree 1, it is sufficient to consider $x,y,z$ on the plane $P:=\{x+y+z=1\}$. Let
$$g(x,y,z):=x+y+z-1$$
be the constraint function. We compute :
$$\mathrm{d}f(x,y,z)=\left(\frac{ax^6+4bx^3y^3}{(ax^3+by^3)^2}-\frac{3bx^2z^4}{(az^3+bx^3)^2}\right)\mathrm{d}x+\left(\frac{ay^6+4by^3z^3}{(ay^3+bz^3)^2}-\frac{3bx^4y^2}{(ax^3+by^3)^2}\right)\mathrm{d}y$$
$$+\left(\frac{az^6+4bx^3z^3}{(az^3+bx^3)^2}-\frac{3by^4z^2}{(ay^3+bz^3)^2}\right)\mathrm{d}z,$$
$$\mathrm{d}g(x,y,z)=\mathrm{d}x+\mathrm{d}y+\mathrm{d}z.$$
Define the $2\times 3$ matrix
$$M:=\begin{pmatrix} \frac{\partial f}{\partial x}(x,y,z) & \frac{\partial f}{\partial y}(x,y,z) & \frac{\partial f}{\partial z}(x,y,z)\\ \frac{\partial g}{\partial x}(x,y,z) & \frac{\partial g}{\partial y}(x,y,z) & \frac{\partial g}{\partial z}(x,y,z) \end{pmatrix}.$$
By Lagrange multipliers theorem, all the 3 sub-determinants of $M$ must vanish at a local minimum $(x,y,z)$ of $f$ on $P$.

Setting
$$A:=ax^3+by^3,\quad B:=az^3+bx^3,\quad ay^3+bz^3,$$
cancelling the 3 sub-determinants of $M$ yields :
\begin{align}
\begin{cases}
B^2C^2(ax^6+4bx^3y^3+3bx^4y^2)-3A^2C^2bx^2z^4-A^2B^2(ay^6+4by^3z^3)&=0\\
B^2C^2(ax^6+4bx^3y^3)-A^2C^2(3bx^2z^4+az^6+4bx^3z^3)+3A^2B^2by^4z^2&=0\\
A^2B^2(ay^6+4by^3z^3+3by^4z^2)-3B^2C^2bx^4y^2-A^2C^2(az^6+4bx^3z^3)&=0\\
x+y+z=1,\ x,y,z>0
\end{cases}.
\end{align}
Labelling the lines $(1)$, $(2)$, $(3)$ and $(4)$, we can see that $(1)-(2)=-(3)$, so that we can forget one of the three first lines.

Here we need to do some (boring) algebra, using the constraints of the fourth line above and maybe some tricks like writing $ax^3=A-by^3$ and $bx^4=(1-y-z)(B-az^3)$. But I am too busy now to try this, and I don’t know if I would try later…

5 people think this answer is useful

Another way.

By C-S $$\sum_{cyc}\frac{x^4}{8x^3+5y^3}=\sum_{cyc}\frac{x^4(3x-y+2z)^2}{(8x^3+5y^3)(3x-y+2z)^2}\geq\frac{\left(\sum\limits_{cyc}(3x^3-x^2y+2x^2z)\right)^2}{\sum\limits_{cyc}(8x^3+5y^3)(3x-y+2z)^2}.$$
Thus, it’s enough to prove that:
$$13\left(\sum\limits_{cyc}(3x^3-x^2y+2x^2z)\right)^2\geq(x+y+z)\sum\limits_{cyc}(8x^3+5y^3)(3x-y+2z)^2.$$
Since the last inequality is cyclic, we can assume that $$x=\min\{x,y,z\}$$.

1. Let $$x\leq z\leq y$$, $$z=x+u$$ and $$y=x+u+v$$.

Thus, $$u$$ and $$v$$ are non-negatives and we need to prove that: $$166(u^2+uv+v^2)x^4+(555u^3+1791u^2v+1454uv^2+109v^3)x^3+$$
$$+(861u^4+3639u^3v+4284u^2v^2+1506uv^3+192v^4)x^2+$$
$$+(555u^5+2474u^4v+3833u^3v^2+2317u^2v^3+153uv^4+166v^5)x+$$
$$+123u^6+547u^5v+1046u^4v^2+843u^3v^3+374u^2v^4+153uv^5+40v^6\geq0,$$ which is obvious;

1. Let $$x\leq y\leq z,$$ $$y=x+u$$ and $$z=x+u+v$$.

Thus, we need to prove that:
$$166(u^2+uv+v^2)x^4+(555u^3-126u^2v-463uv^2+109v^3)x^3+$$
$$+(861u^4-195u^3v-1467u^2v^2-411uv^3+192v^4)x^2+$$
$$+(555u^5+301u^4v-513u^3v^2-112u^2v^3+479uv^4+166v^5)x+$$
$$+123u^6+191u^5v+156u^4v^2+331u^3v^3+496u^2v^4+253uv^5+40v^6\geq0.$$
Easy to show that:
$$166(u^2+uv+v^2)\geq498uv,$$
$$555u^3-126u^2v-463uv^2+109v^3\geq-249\sqrt{u^3v^3},$$
$$861u^4-195u^3v-1467u^2v^2-411uv^3+192v^4\geq-1494u^2v^2,$$
$$555u^5+301u^4v-513u^3v^2-112u^2v^3+479uv^4+166v^5\geq747\sqrt{u^5v^5}$$ and
$$123u^6+191u^5v+156u^4v^2+331u^3v^3+496u^2v^4+253uv^5+40v^6\geq1494u^3v^3.$$
Thus, after substitution $$x=t\sqrt{uv}$$ it’s enough to prove that
$$498t^4-249t^3-1494t^2+747t+1494\geq0,$$ which is true because
$$498t^4-249t^3-1494t^2+747t+1494=$$
$$=249(t+1)(2t^3-3t^2-3t+6)=249(t+1)(t^3+2-3t+t^3+4-3t^2)\geq$$
$$\geq249(t+1)\left(3\sqrt{t^3\cdot1^2}-3t+3\sqrt{\left(\frac{t^3}{2}\right)^2\cdot4}-3t^2\right)=0.$$
Done!

3 people think this answer is useful

For checking purposes.

Making $$y = \lambda, \ z = \mu x$$ and substituting into

$$f(x,y,z) = \frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} – \frac{x+y+z}{13}$$

giving

$$g(x,\lambda,\mu) =x\left( \frac{1}{5 \lambda ^3+8}+\frac{\lambda ^4}{8 \lambda ^3+5 \mu ^3}+\frac{\mu ^4}{8 \mu ^3+5}-\frac{1}{13} (\lambda +\mu +1)\right)$$

and discarding $$x > 0$$ we get

$$\mathcal{G}(\lambda,\mu) = \frac{1}{5 \lambda ^3+8}+\frac{\lambda ^4}{8 \lambda ^3+5 \mu ^3}+\frac{\mu ^4}{8 \mu ^3+5}-\frac{1}{13} (\lambda +\mu +1)$$

Now solving the stationary conditions

$$\nabla\mathcal{G}(\lambda,\mu) = 0$$

we have the feasible stationary points with qualification.

$$\left[ \begin{array}{cccl} \lambda & \mu & \mathcal{G}(\lambda,\mu) & \mbox{kind} \\ 1. & 1. & 0. & \mbox{min} \\ 0.485435 & 0.715221 & 0.000622453 & \mbox{min}\\ 0.646265 & 0.811309 & 0.000758688 & \mbox{saddle} \\ 1.37554 & 0.688678 & 0.000863479 & \mbox{min} \\ 1.25 & 0.77611 & 0.000941355 & \mbox{saddle} \\ 1.38778 & 1.85522 & 0.00123052 & \mbox{min} \\ 1.34211 & 1.74761 & 0.00123288 & \mbox{saddle} \\ \end{array} \right]$$

so the best solution is at $$x = y = z = 1$$

Attached the level contours for $$\mathcal{G}(\lambda,\mu)$$ with the stationary points in red. 2 people think this answer is useful

Not sure, if I missed out anything here. Take a look.

For non negative, $X,Y,Z$,
We can perhaps use Titu’s inequality (a mix of Holder and CS), sometimes called Titu’s screw lemma (https://en.wikipedia.org/wiki/Nesbitt%27s_inequality).
\begin{equation}
\sum_{k=1}^{n}{\frac{x_{k}^{2}}{a_{k}}} \ge \frac{\left(\sum_{k=1}^{n}{x_{k}}\right)^{2}}{\sum_{k=1}^{n}{a_{k}}}
\end{equation}

With $n\to3$ terms, $x_{1}\to X^{2},x_{2} \to Y^{2}, x_{3} \to Z^{2}$ and $a_{1} \to A, a_{2}\to B, a_{3} \to C$, we will have

\begin{eqnarray*}
\frac{\left(X^2\right)^{2}}{A}+\frac{\left(Y^2\right)^{2}}{B}+\frac{\left(Z^2\right)^{2}}{C} &\ge& \frac{\left(X^{2}+Y^{2}+Z^{2}\right)^{2}}{A+B+C} \\
\end{eqnarray*}

With
\begin{eqnarray*}
A &=& \alpha X^{3} +\beta Y^{3} \\
B &=& \alpha Y^{3} +\beta Z^{3} \\
C &=& \alpha Z^{3} +\beta X^{3}
\end{eqnarray*}

where,
\begin{eqnarray*}
A+B+C &=& (\alpha+\beta) \left(X^{3} + Y^3+Z^3 \right)
\end{eqnarray*}

\begin{eqnarray}
\frac{X^4}{A}+\frac{Y^4}{B}+\frac{Z^4}{C} &=&\frac{\left(X^2\right)^{2}}{A}+\frac{\left(Y^2\right)^{2}}{B}+\frac{\left(Z^2\right)^{2}}{C}\\
&\ge& \frac{\left(X^{2}+Y^{2}+Z^{2}\right)^{2}}{A+B+C} \\
&=& \frac{\left(X^{2}+Y^{2}+Z^{2}\right)^{2}}{(\alpha+\beta) \left(X^{3} + Y^3+Z^3 \right)} \\
&\overset{(p)}{\ge}& \frac{\left(X^{3}+Y^{3}+Z^{3}\right)\left(X+Y+Z\right)}{(\alpha+\beta) \left(X^{3} + Y^3+Z^3 \right)} \\
&=& \frac{\left(X+Y+Z\right)}{(\alpha+\beta)}
\end{eqnarray}

QED.

Here $(p)$ is from the fact that,

\begin{eqnarray*}
(X^2+Y^2+Z^2)^{2} -\left(X^{3}+Y^{3}+Z^{3} \right) (X+Y+Z) &=& XY(X-Y)^{2}+YZ(Y-Z)^{2}+ZX(Z-X)^{2} \\
&\ge& 0
\end{eqnarray*}

Here $\alpha=8$ and $\beta=5$.

2 people think this answer is useful

Let’s reform this inequality in a way such that we can comprehend it better. Define $a=\dfrac{y}{x}$ and $b=\dfrac{z}{y}$, therefore $\dfrac{x}{z}={1\over ab}$. We can suume without lose of generality that $a,b\le1$ We need to prove that $$\dfrac{x}{8+5\left(\dfrac{y}{x}\right)^3}+\dfrac{y}{8+5\left(\dfrac{z}{y}\right)^3}+\dfrac{z}{8+5\left(\dfrac{x}{z}\right)^3}\ge\dfrac{x+y+z}{13}$$by dividing the two sides of the inequality by $x$ and substituting $a,b,c$ we have that$$\dfrac{1}{8+5\left(\dfrac{y}{x}\right)^3}+\dfrac{\dfrac{y}{x}}{8+5\left(\dfrac{z}{y}\right)^3}+\dfrac{\dfrac{z}{y}}{8+5\left(\dfrac{x}{z}\right)^3}\ge\dfrac{1+\dfrac{y}{x}+\dfrac{z}{x}}{13}$$and $$\dfrac{1}{8+5a^3}+\dfrac{a}{8+5b^3}+\dfrac{a^4b^4}{5+8a^3b^3}\ge \dfrac{1}{13}+\dfrac{a}{13}+\dfrac{ab}{13}$$which is equivalent to $$\left(\dfrac{1}{8+5a^3}-\dfrac{1}{13}\right)+\left(\dfrac{a}{8+5b^3}-\dfrac{a}{13}\right)+\left(\dfrac{a^4b^4}{5+8a^3b^3}-\dfrac{ab}{13}\right)\ge 0$$by simplifying each of the components and multiplying both sides in $\dfrac{13}{5}$ we obtain$$\dfrac{1-a^3}{8+5a^3}+\dfrac{a(1-b^3)}{8+5b^3}+\dfrac{a^4b^4-ab}{5+8a^3b^3}\ge0$$below is a depiction of $f(a,b)=\dfrac{1-a^3}{8+5a^3}+\dfrac{a(1-b^3)}{8+5b^3}+\dfrac{a^4b^4-ab}{5+8a^3b^3}$ for $0\le a,b\le 1$ which proves the inequality graphically (I believe that Lagrange multipliers or any other method based on 1st order derivations may help but i hadn’t much time to think on it hope you find an analytic way) but neither such a time i spent on the problem nor a computer is given us in the exam 🙂 also i appreciate if any one updates his/her comment with such an analytical method. I’m really curious about that…..

2 people think this answer is useful

$$\color{green}{\textbf{Light version (24.01.20).}}$$

$$\color{brown}{\textbf{Inequalities for cubic root.}}$$

Searching of the polynomials in the forms of
$$\begin{cases} P_4(s)=s(1+as^3) – (1+a)s^3 = as^4 -(a+1)s^3+s\\[4pt] P_7(s)=(5+8s^3)(1-b+bs^3) – s(13-c+cs^3)\\ \qquad = 8bs^6-cs^4+(8-3b)s^3+(c-13)s+5-5b \end{cases}$$
under the conditions
$$P’_4(1)=P’_7(1)=P”_7(1) = 0,$$
allows to obtain the coefficients $$a,b,c:$$
$$\begin{cases} 4a-3(1+a)+1=0\\ 48b-4c+3(8-3b)+c-13=0\\ 240b-12c+6(8-3b)=0 \end{cases} \Rightarrow \begin{cases} a = 2\\ 39b-3c = -11\\ 222b-12c = -48, \end{cases}$$
$$a=2,\quad b=-\dfrac2{33},\quad c=\dfrac{95}{33},$$

then
$$\begin{cases} P_4(s) = s(1+2s^3) – 3s^3 = s(1-s)^2(2s+1)\\ 33P_7(b)=(35-2s^3)(5+8s^3) – s(334+95s^3) = (1-s)^3(16s^3+48s^2+191s+175). \end{cases}$$
If $$s\in[0,1]\$$ then $$P_4(s)\ge0,\ P_7(s)\ge0.$$

Applying of the substitution $$s=\sqrt[\large 3]{1-t\large\mathstrut}\$$

$$\dfrac{(13-8t)(33+2t)}{429-95t} \ge \sqrt[\large3]{1-t\mathstrut} \ge \dfrac{3(1-t)}{3-2t}\quad\text{if} \quad t\in[0,1]\tag1$$ On the other hand, the function
$$S(t)=\sqrt[\Large3]{\dfrac{5t\mathstrut}{13-8t}},\quad t\in[0,1]$$

has the inverse one in the form of
$$T(s)=\dfrac{13s^3}{8s^3+5},\quad s\in[0,1].$$

If $$s=S(t),$$ then
\begin{align} &\dfrac{15+11t-11t^2}{3(13-8t)}-S(t) = \dfrac{15+11T(s)-11T^2(s)}{3(13-8T(s))}-s\\[8pt] & = \dfrac{49s^6-312s^4+383s^3-195s+75}{312s^2+195} = \dfrac{(s+1)^2(7s+5)(7s^3+9s^2-30s+15)}{39(8s^2+5)},\\[8pt] &7s^3+9s^2-30s+15 = 7(1-s)(1-s^2)+8(1-s)(2-s)+s, \end{align}

$$S(t) = \sqrt[\Large3]{\dfrac{5t\mathstrut}{13-8t}} \le \dfrac{15+11t-11t^2}{3(13-8t)},\quad t\in[0,1].\tag2$$ $$\color{brown}{\textbf{Primary transformations.}}$$

The given inequality WLOG can be presented in the forms of
$$x\ge y,\quad x\ge z,\quad \dfrac{x^4}{8x^3+5y^3}+\dfrac{y^4}{8y^3+5z^3}+\dfrac{z^4}{8z^3+5x^3} \ge \dfrac1{13}(x+y+z),\tag3$$

or
$$\dfrac{13x^4}{8x^3+5y^3}-x + \dfrac{13y^4}{8y^3+5z^3}-y + \dfrac{13z^4}{8z^3+5x^3}-z \ge 0,$$

$$\dfrac{x^3-y^3}{8x^3+5y^3} + \dfrac yx\,\dfrac{y^3-z^3}{8y^3+5z^3} – \dfrac zx\,\dfrac{x^3-z^3}{5x^3+8z^3} \ge 0.\tag4$$

$$\color{brown}{\mathbf{Case\ \ z < y \le x.}}$$

Taking in account $$(1),$$ inequality $$(4)$$ in the notation
$$\dfrac{z^3}{x^3} = 1-u,\quad \dfrac{y^3}{x^3} = 1-uv,\quad (u,v)\in[0,1]^2, \tag5$$
\begin{align} &\dfrac{x^3-y^3}{8x^3+5y^3} = \dfrac{uv}{8+5(1-uv)},\quad \dfrac yx = \sqrt[\large3]{1-uv\mathstrut},\\[8pt] &\dfrac{y^3-z^3}{8y^3+5z^3} = \dfrac{u-uv}{8(1-uv)+5(1-u)},\\[8pt] &\dfrac{x^3-z^3}{5x^3+8z^3} = \dfrac{u}{5+8(1-u)},\quad \dfrac zx = \sqrt[\large3]{1-u\mathstrut}, \end{align}

takes the form of $$f_1(u,v) \ge 0,$$ where
\begin{align} &f_1(u,v) = u\left(\dfrac{v}{13-5uv} + \dfrac{3(1-uv)}{3-2uv}\,\dfrac{1-v}{13-5u-8uv} – \dfrac{33+2u}{429-95u}\right)\\[8pt] & = \dfrac{u^2(A(u)+vB(u)+v^2C(u)+v^3D(u))}{(3-2uv)(13-5u-8uv)(13-5uv)(429-95u)}, \end{align}
\begin{align} & A(u) = 1716+390u,\\ & B(u) = -1716+1480u-410u^2,\\ & C(u) = 1716-4769u-1641u^2+100u^3,\\ & D(u) = 429u + 2545u^2+160u^3,\\ & A(u)+vB(u)+v^2C(u)+v^3D(u) = (1-v)(1-v^2)A(u)+v(1-v)^2(A(u)+B(u))\\ & +v^2(1-v)(3A(u)+2B(u)+C(u))+v^3(A(u)+B(u)+C(u)+D(u))\\ & = (1-v)(1-v^2)(1716+390u)+v(1-v)^2(1870u-410u^2)\\ & +v^2(1-v)(3432-639u-2461u^2+100u^3)+26v^3(1-u)(66-29u-10u^2) \ge 0 \end{align} $$\!\mathstrut^{\phantom{\dfrac\mathstrut\mathstrut}^{\LARGE=}}$$ Therefore, $$f_1(u,v)\ge0.$$

The case is proved.

$$\color{brown}{\mathbf{Case\ \ y \le z \le x.\ Additional\ transformations.}}$$

Using the notation
$$\dfrac{5(x^3-z^3)}{5x^3+8z^3} = 1-u,\quad \dfrac{5(z^3-y^3)}{5z^3+8y^3} = 1-v,\quad (u,v)\in[0,1]^2, \tag6$$

one can get
$$\dfrac{z^3}{x^3} = \dfrac{5u}{13-8u},\quad\dfrac{y^3}{z^3} = \dfrac{5v}{13-8v},\quad \dfrac{y^3}{x^3} = \dfrac{25uv}{(13-8u)(13-8v)},$$
$$\dfrac{x^3-y^3}{8x^3+5y^3} = \dfrac{(13-8u)(13-8v)-25uv}{8(13-8u)(13-8v)+125uv} = \dfrac{13-8(u+v)+3uv}{104-64(u+v)+49uv}.\tag7$$

$$\color{brown}{\mathbf{Case\ \ y \le z \le x,\ u+v \ge \dfrac{13}8.}}$$

Taking in account $$(2),$$ the inequality $$(4)$$ takes the stronger form of $$f_2(u,v)\ge0,$$ where
\begin{align} &f_2(u,v) = 5\dfrac{13-8(u+v)+3uv}{104-64(u+v)+49uv}- (1-v)S(u)S(v) – (1-u)S(u)\\[8pt] & \ge 5\dfrac{13-8(u+v)+3uv}{104-64(u+v)+49uv} – \dfrac{15+11u-11u^2}{3(13-8u)} \left((1-v)\dfrac{15+11v-11v^2}{3(13-8v)}+1-u\right)\\ & = \dfrac{g_2(u,v)}{9(104-64(u+v)+49uv)(13-8u)(13-8v)},\\[8pt] &g_2(u,v)=5(13-8(u+v)+3uv)(39-24u)(39-24v)\\[4pt] &-((1-v)(15+11(1-v)v)+(39-24v)(1-u))\\[4pt] &\times(15+11(1-u)u)(104-64(u+v)+49uv). \end{align}

Let $$p=1-u,\ \ q=1-v,$$ then $$p+q \in \left[0,\dfrac58\right],$$

\begin{align} &g^\,_2(p,q) = 5(5(p+q)+3pq)(15+24p)(15+24q)\\[4pt] &-(q(15+11(1-q)q)+(15+24q)p)(15+11(1-p)p)(25+15(p+q)+49pq)\\[4pt] &= 1500p^2+1500pq+1500q^2\\[4pt] &+1650p^3-4050p^2q-4600pq^2+1650q^3\\[4pt] &+2475p^4-495p^3q-17360p^2q^2-4400pq^3+2475q^4\\[4pt] &+12045p^4q+924p^3q^2-5324p^2q^3+9900pq^4\\[4pt] &+12936p^4q^2+4114p^3q^3+4114p^2q^4-5929p^3q^4\\[4pt] \end{align}

Since
$$pq \le \dfrac14(p+q)^2,\quad p^3-p^2q-pq^2+q^3 = (p-q)(p^2-q^2) \ge 0,$$

then
\begin{align} &g^\,_2(p,q) \ge 375(4(p+q)^2-4pq)\\[4pt] & + 1650(p-q)(p^2-q^2) – 3000pq(p+q)\\[4pt] &+2475(p^2-q^2)^2 -pq(495p^2+12410pq+4400q^2)\\[4pt] &+9900pq(p-q)(p^2-q^2)\\[4pt] &+4114p^2q^3(p(1-p)+ q(1-q)+p^2+q^2-2pq)\\[4pt] &\ge 1125(p+q)^2-3000pq(p+q)-6208pq(p+q)^2 + 0 + 0\\[4pt] &\ge 1125(p+q)^2-750(p+q)^3-1552(p+q)^4\\[4pt] &\ge \left(1125 – 750\cdot\dfrac58-1552\cdot\dfrac{25}{64}\right)(p+q)^2 \ge 0. \end{align}

The case is proved.

$$\color{brown}{\mathbf{Case\ \ y \le z \le x,\ u+v \le \dfrac{13}8.}}$$

From $$(7)$$ should
\begin{align} &\dfrac{(49\, \dfrac{x^3-y^3}{8x^3+5y^3}-3)}{100} = \dfrac{13-8(u+v)}{416-256(u+v)+49(2\sqrt{uv})^2} \ge \dfrac{13-8(u+v)}{416-256(u+v)+49(u+v)^2}. \end{align}

Since
$$\dfrac1{49}\left(100\dfrac{13-8t}{416-256t+49t^2}+3\right) = \dfrac{(2-t)(26-3t)}{416-256t+49t^2}$$

and
$$\dfrac{26-3t}{416-256t+49t^2} – \dfrac1{800}(50+21t+17t^2) = \dfrac{t(2-t)(833t^2-1657t+832)}{800(49t^2-256t+416)}$$ then
$$\dfrac{x^3-y^3}{8x^3+5y^3}\ge R(u+v),$$
where

$$R(t) = \dfrac1{800}(2-t)(50+21t+17t^2),\quad t\in[0,2].\tag8$$

Therefore, the inequality $$(3)$$ takes the stronger form of $$f_3(u,v)\ge0,$$ where
\begin{align} &f_3(u,v) = 5R(u+v)- (1-v)S(u)S(v) – (1-u)S(u)\\[8pt] & \ge \dfrac{2-u-v}{160}(50+21(u+v)+17(u+v)^2)\\[8pt] & – \dfrac{15+11u-11u^2}{3(13-8u)} \left((1-v)\dfrac{15+11v-11v^2}{3(13-8v)}+1-u\right)\\ & = \dfrac{g_3(u,v)}{1440(13-8u)(13-8v)},\\[8pt] \end{align}

where
\begin{align} &g^\,_3(u,v) = (50+21(u+v)+17(u+v)^2)(2-u-v)(39-24u)(39-24v)\\[4pt] &-160((1-v)(15+11(1-v)v)+(39-24v)(1-u))(15+11(1-u)u),\\[4pt] &g^\,_3(1-u,1-v) = (160-89(u+v)+17(u+v)^2)(u+v)(15+24u)(15+24v)\\[4pt] & – 160(15+11(1-u)u)((15+11(1-v)v)v+u(15+24v))\\[4pt] &= 11175u^2-1815u^3+6120u^4-8850uv-8325u^2v+15456u^3v+9792u^4v\\[4pt] &+11175v^2-11845uv^2-46448u^2v^2+29376u^3v^2\\[4pt] &-1815v^3-7424uv^3+10016u^2v^3+6120v^4+9792uv^4 \end{align}

In the matrix form,
$$g^\,_3(1-u,1-v) = \mu(u,v,G_3) = \begin{pmatrix} 1 \\ v \\ v^2 \\ v^3 \\ v^4 \end{pmatrix}^T G_3 \begin{pmatrix} 1 \\ u \\ u^2 \\ u^3 \\ u^4 \end{pmatrix},\tag9$$

where
$$G_3 = \begin{pmatrix} 0 & 0 & 11175 & -1815 & 6120 \\ 0 & -8850 & -8325 & 15456 & 9792 \\ 11175 & -11845 & -46448 & 29376 & 0 \\ -1815 & -7424 & 10016 & 0 & 0 \\ 6120 & 9792 & 0 & 0 & 0 \end{pmatrix}.\tag{10}$$

At the same time:

• $$(u-v)^2(1-u-v)^2 = \begin{pmatrix} 1 \\ v \\ v^2 \\ v^3 \\ v^4 \end{pmatrix}^T \begin{pmatrix} 0 & 0 & 1 & -2 & 1 \\ 0 &-2 & 2 & 0 & 0 \\ 1 & 2 & -2 & 0 & 0 \\ -2 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ \end{pmatrix} \begin{pmatrix} 1 \\ u \\ u^2 \\ u^3 \\ u^4 \end{pmatrix},$$

• $$g_3(u,v) = 6120(u-v)^2(1-u-v)^2 + uv(9792(u-v)(u^2-v^2)+15456(u-v)^2)\\ + g^\,_{32}(u,v) = g^\,_{30}(u,v) + g^\,_{31}(u,v) + g^\,_{32}(u,v) = \mu(u,v,G_{30}+G_{31}+G_{32}),$$
where
$$G_{30} = \begin{pmatrix} 0 & 0 & 6120 & -12240 & 6120 \\ 0 & -12240 & 12240 & 0 & 0 \\ 6120 & 12240 & -12240 & 0 & 0 \\ -12240 & 0 & 0 & 0 & 0 \\ 6120 & 0 & 0 & 0 & 0 \end{pmatrix},$$
$$G_{31} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 15456 & 9792 \\ 0 & 0 & -30912 & -9792 & 0 \\ 0 & -15456 & -9792 & 0 & 0 \\ 0 & 9792 & 0 & 0 & 0 \end{pmatrix},$$
$$G_{32} = \begin{pmatrix} 0 & 0 & 5055 & 10425 & 0 \\ 0 & 3390 & -3915 & 0 & 0 \\ 5055 & -24085 & -3296 & 39168 & 0 \\ 10425 & -22880 & 19808 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix},$$$$g^\,_{30}(u,v) \ge 0,\quad g^\,_{31}(u,v) \ge 0.$$

Since $$\!\mathstrut^{\phantom{\dfrac\mathstrut\mathstrut}^{\LARGE=}}$$ $$\!\mathstrut^{\phantom{\dfrac\mathstrut\mathstrut}^{\LARGE,}}$$

then, similarly to the first case,
$$g^\,_{32}(u,v)= \begin{pmatrix} 1 \\ v \\ v^2 \\ v^3 \end{pmatrix}^T \begin{pmatrix} 0 & 0 & 5055 & 10425 \\ 0 & 3390 & -3915 & 0 \\ 5055 & -24085 & -3296 & 39168 \\ 10425 & -22880 & 19808 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ u \\ u^2 \\ u^3 \end{pmatrix}\\ =\begin{pmatrix} (1-v)(1-v^2) \\ v(1-v)^2 \\ v^2(1-v) \\ v^3 \end{pmatrix}^T \begin{pmatrix} 0 & 0 & 5055 & 10425 \\ 0 & 3390 & 1140 & 10425 \\ 5055 & -17305 & 40039 & 70443 \\ 15480 & -43575 & 17652 & 49593 \end{pmatrix} \begin{pmatrix} 1 \\ u \\ u^2 \\ u^3 \end{pmatrix},$$
wherein

$$5055 – 17305u + 4039u^2 + 70443u^3 = 5055(1-2u)^2 + u(2915-16181u+70443) \ge 0,$$
$$15480 – 43575u + 17652u^2 + 49593u^3 = 15480(1-2u)^2 +3u(6115 -14756u + 16531u^2) \\ \ge0,$$ Thus, $$g^\,_{32}(u,v)\ge 0$$ and $$g_3(u,v) \ge 0.$$

PROVED.

1 people think this answer is useful

This is too long to fit into a comment. I wanted to ask a question about my proof on this problem. (It might help discover another proof)

This proof has a flaw — From $AB \ge C$ and $A \ge D$, I wrongly implied that $DB \ge C$.

Is there a way to slightly modify it such that it can prove the statement or is it completely wrong?

Seeing that the inequality is homogeneous (meaning that the transformation
$(x, y, z) \mapsto (kx, ky, kz)$ does not change anything), it is natural to impose a constraint on it. So let us assume
without the loss of generality that $xyz=1$.

From Cauchy-Schwarz Inequality,

$$([8x^3+5y^3]+[8y^3+5z^3]+[8z^3+5x^3])(\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3})\geqslant (x^2+y^2+z^2)^2$$

Since (By AM-GM) $$[8x^3+5y^3]+[8y^3+5z^3]+[8z^3+5x^3] = 13(x^3+y^3+z^3) \geqslant 13(3 \sqrt{(xyz)^3}) = 13(3)$$

Therefore

$([8x^3+5y^3]+[8y^3+5z^3]+[8z^3+5x^3])(\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3}) \geqslant (13)(3)(\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3}) \geqslant (x^2+y^2+z^2)^2$

Therefore

$$\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} \geqslant \frac{(x^2+y^2+z^2)^2}{(13)(3)}$$

Now it reamins to prove that $\frac{(x^2+y^2+z^2)^2}{(13)(3)} \geqslant \frac{x+y+z}{13}$, i.e.

$$(x^2+y^2+z^2)(x^2+y^2+z^2)\geqslant 3(x+y+z)$$

which is straightforward by AM-GM:

Notice that for all $xyz=1$

$$(x – 1)^2 + (y-1)^2 + (z – 1)^2 \ge 0$$
$$x^2 + y^2 + z^2 – 2a – 2b – 2c + 3 \ge 0$$
$$x^2 + y^2 + z^2 \ge -3 + (x + y + z) + (x + y + z)$$

But by AM-GM, $x + y + z \ge 3\sqrt{xyz} = 3$. So,
$$x^2 + y^2 + z^2 \ge -3 + 3 + (x + y + z)$$
$$x^2 + y^2 + z^2 \ge x + y + z \ge 3$$

1 people think this answer is useful

I have finally found a solution . In fact we start to study the 2 variables version of this inequality we have :

$$\frac{a^4}{8a^3+5b^3}+\frac{b^4}{8b^3+5a^3}\geq \frac{a+b}{13}$$

Proof:

We have with $$x=\frac{a}{b}$$ :
$$\frac{x^4}{8x^3+5}+\frac{1}{8+5x^3}\geq \frac{1+x}{13}$$
Or
$$\frac{5}{13}(x – 1)^2 (x + 1) (x^2 + x + 1) (5 x^2 – 8 x + 5)\geq 0$$

So we have (if we permute the variables $$a,b,c$$ and add the three inequalities ) :

$$\sum_{cyc}\frac{a^4}{8a^3+5b^3}+\sum_{cyc}\frac{a^4}{8a^3+5c^3}\geq \frac{a+b+c}{6.5}$$

If we have $$\sum_{cyc}\frac{a^4}{8a^3+5b^3}\geq\sum_{cyc}\frac{a^4}{8a^3+5c^3}$$

We have :
$$\sum_{cyc}\frac{a^4}{8a^3+5b^3}\geq \frac{a+b+c}{13}$$
But also
$$\frac{(a-\epsilon)^4}{8(a-\epsilon)^3+5b^3}+\frac{(b)^4}{8(b)^3+5(c+\epsilon)^3}+\frac{(c+\epsilon)^4}{8(c+\epsilon)^3+5(a-\epsilon)^3}\geq \frac{a+b+c}{13}$$
If we put $$a\geq c$$ and $$\epsilon=a-c$$

We finally obtain :
$$\sum_{cyc}\frac{a^4}{8a^3+5c^3}\geq \frac{a+b+c}{13}$$

If we have $$\sum_{cyc}\frac{a^4}{8a^3+5b^3}\leq\sum_{cyc}\frac{a^4}{8a^3+5c^3}$$

The proof is the same as above .

So all the cases are present so it’s proved !

0 people think this answer is useful

We want to prove the inequality
$$\sum_{cyc}\frac{x^4}{8x^3+5y^3}\geq \frac{\sum_{cyc}x}{13}\tag 1$$
where $$x,y,z>0$$.

Set
$$\Pi(x,y,z):=\sum_{cyc}\frac{x^4}{8x^3+5y^3}-\frac{x+y+z}{13}$$
Then easily
$$\Pi(x,y,z)= \sum_{cyc}\left(x\frac{x^3}{8x^3+5y^3}-\frac{x}{13}\right)=\sum_{cyc}x\frac{13x^3-8x^3-5y^3}{13(8x^3+5y^3)}=$$
$$=\frac{5}{13}\sum_{cyc}x\frac{x^3-y^3}{8x^3+5y^3}.$$
Also (ecxept from the symetry) $$\Pi(x,y,z)$$ is homogeneous of degree 1. Hence $$\Pi(\lambda x,\lambda y,\lambda z)=\lambda \Pi(x,y,z)$$.

The case $$x\geq y\geq z$$

If $$x\geq y\geq z$$, we can set $$\lambda=1/z$$, then
$$\Pi(x,y,z)=\lambda^{-1}\Pi(\lambda x,\lambda y,\lambda z)=z^{-1}\Pi\left(\frac{x}{z},\frac{y}{z},1\right).$$
Set $$a=\frac{x}{z}$$, $$b=\frac{y}{z}$$, then also $$a\geq b\geq 1$$ and
$$\Pi(x,y,z)=z\Pi(a,b,1)=$$
$$=z\left(\frac{1-a^3}{8+5a^3}+\frac{b(-1+b^3)}{5+8b^3}+\frac{a^4-ab^3}{8a^3+5b^3}\right)\textrm{, }a\geq b\geq 1\tag 2$$
Set now
$$f(x,y)=\frac{x}{8x^3+5 y^3}.$$
Then (2) becomes
$$a^3\left(\frac{a}{8a^3+5b^3}-\frac{1}{8+5a^3}\right)+b^3\left(\frac{b}{5+8b^3}-\frac{a}{8a^3+5b^3}\right)-$$
$$-\left(\frac{b}{8b^3+5}-\frac{1}{8+5a^3}\right)\geq 0$$
Hence equivalent we must show that
$$a^3(f(a,b)-f(1,a))+b^3(f(b,1)-f(a,b))-(f(b,1)-f(1,a))\geq 0\Leftrightarrow$$
$$a^3f(a,b)+f(1,a)(1-a^3)\geq b^3f(a,b)+f(b,1)(1-b^3)\Leftrightarrow$$
$$f(1,a)(a^3-1)-f(b,1)(b^3-1)\leq (a^3-b^3)f(a,b)\tag 3$$
But when $$a\geq b\geq 1$$, setting $$k=a^3-1$$ and $$l=b^3-1$$, we have $$k\geq l\geq 0$$. Then (3) becomes
$$k\frac{1}{5a^3+8}-l\frac{b}{8b^3+5}\leq (k-l)\frac{a}{8a^3+5b^3}\Leftrightarrow$$
$$\frac{k}{5k+13}-\frac{lb}{8l+13}\leq (k-l)\frac{a}{8k+5l-13}\Leftrightarrow$$
$$\frac{k}{k-l}\frac{1}{x_1}-\frac{l}{k-l}\frac{b}{y_1}\leq \frac{a}{x_1+y_1-13}$$
Or equivalent (we set $$x_1=5k+13$$, $$y_1=8l+13$$) if:
$$f_0(x_1,y_1):=ky_1^2-blx_1^2+x_1y_1(-bl+k-ak+al)-13ky_1+13lb x_1\leq 0.$$
This last inequality is true, as one can see using $$1\leq b\leq a$$ and $$k\geq l\geq 0$$, $$k=a^3-1$$, $$l=b^3-1$$, $$x_1\geq 13$$, $$y_1\geq 13$$. Actualy it holds
$$f_0(x_1,y_1)>f_0(x_0,y_0)\textrm{, }\forall (x_1,y_1)\in D=(x_0,+\infty)\times(y_0,+\infty)$$
where
$$x_0=\frac{13k(k-ak+(a+b)l)}{(a-1)^2k^2+2(b+a(1-a+b))kl+(a-b)^2l^2}$$
$$y_0=\frac{13bl(k+ak+(-a+b)l)}{(a-1)^2k^2+2(b+a(1-a+b))kl+(a-b)^2l^2}$$
are the points such that $$\partial_xf_0(x_0,y_0)=\partial_yf_0(x_0,y_0)=0$$ and
$$f_0(x_0,y_0)=\frac{169abkl(l-k)}{(a-1)^2k^2+2(b+a(1-a+b))kl+(a-b)^2l^2}\leq 0$$
and $$(a-1)^2k^2+2(b+a(1-a+b))kl+(a-b)^2l^2>0$$, $$a>1$$, $$a\geq b\geq 1$$. But when $$x_1\geq 13$$ and $$y_1\geq 13$$, we have $$f_0(x_1,y_1)\leq 0$$.

About the case $$z\geq y\geq x$$

If $$z\geq y\geq x$$, then we have $$\Pi(\lambda x,\lambda y,\lambda z)=\lambda \Pi(x,y,z)$$. Set $$\lambda=1/x$$ and $$a=z/x$$, $$b=y/x$$, $$a\geq b\geq 1$$. Then
$$\Pi(x,y,z)=x^{-1}\Pi(1,b,a)=$$
$$b^3\left(\frac{b}{8b^3+5a^3}-\frac{1}{8+5b^3}\right)+a^3\left(\frac{a}{5+8a^3}-\frac{b}{8b^3+5a^3}\right)-$$
$$-\left(\frac{a}{8a^3+5}-\frac{1}{8+5b^3}\right)\geq 0\tag 5$$
Set
$$f(x,y)=\frac{x}{8x^3+5y^3},$$
then (5) becomes
$$b^3(f(b,a)-f(1,b))+a^3(f(a,1)-f(b,a))-f(a,1)+f(1,b)\geq 0\Leftrightarrow$$
$$(a^3-1)f(a,1)-(b^3-1)f(1,b)\geq (a^3-b^3)f(b,a).$$
Set now $$k=a^3-1$$ and $$l=b^3-1$$. Then $$k\geq l\geq 0$$. Then
$$k\frac{a}{8 a^3+5}-l\frac{1}{8+5 b^3}\geq (k-l)\frac{b}{8b^3+5a^3}\Leftrightarrow$$
$$\frac{ka}{8k+13}-\frac{l}{5 l+13}\geq (k-l)\frac{b}{8k+5l+13}\Leftrightarrow$$
$$\frac{ka}{k-l}\frac{1}{x_1}-\frac{l}{k-l}\frac{1}{y_1}\geq \frac{b}{x_1+y_1-13}$$
Or equivalent ($$x_1=8k+13$$, $$y_1=5l+13$$):
$$f_0(x_1,y_1):=aky_1^2-lx_1^2+x_1y_1(bl-l-bk+ak)-13aky_1+13l x_1\geq 0.$$
This last inequality is true and as one can see using $$1\leq b\leq a$$ and $$k\geq l\geq 0$$, $$k=a^3-1$$, $$l=b^3-1$$, $$x_1\geq 13$$, $$y_1\geq 13$$. Actualy holds
$$f_0(x_1,y_1)>f_0(x_0,y_0)\textrm{, }\forall (x_1,y_1)\in D=(x_0,+\infty)\times(y_0,+\infty)$$
where
$$x_0=\frac{13ak(ak+l+b(l-k))}{a^2k^2+(b(k-l)+l)^2+2ak(l+b(l-k))}$$
$$y_0=\frac{13l((a+b)k+l-lb)}{a^2k^2+(b(k-l)+l)^2+2ak(l+b(l-k))}$$
are the points such that $$\partial_xf(x_0,y_0)=\partial_yf_0(x_0,y_0)=0$$ and
$$f_0(x_0,y_0)=\frac{169abkl(l-k)}{a^2k^2+(b(k-l)+l)^2+2ak(l+b(l-k))}\leq 0$$
and $$a^2k^2+(b(k-l)+l)^2+2ak(l+b(l-k))>0$$, $$a>1$$, $$a\geq b\geq 1$$. But when $$x_1\geq 13$$ and $$y_1\geq 13$$, we have $$f_0(x_1,y_1)\geq 0$$. QED