Jtdylktuy

This is a question of the symmetric type, such as listed in:

• 
  Why does Group Theory not come in here?
  

• 
  How prove this inequality $(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) leq 64$
  

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

 0.00000000000000E+0000 < f < 4.80709198767699E-0002

This inequality is very strong and I am sure that it's impossible to prove it during competition without calculator.

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

By the way, the Buffalo Way helps:

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

Hence, $frac{13}{5}prodlimits_{cyc}(8x^3+5y^3)left(sumlimits_{cyc}frac{x^4}{8x^3+5y^3}-frac{x+y+z}{13}right)=$

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

Done!

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
$$frac{(x^2+y^2+z^2)^2}{13(x^3+y^3+z^3)}geqfrac{x+y+z}{13}$$

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.

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 `w`and 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) 

read and interpret file: #pD:/maxima/ineq1775572.mac

(%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 `w`and 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 `w`and 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.

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

To proof that this function is larger than
$$left(156,t^8+531,t^7+2,t^6-632,t^5-152,t^4+867,t^3+834,t^2+
299,t+40right),u^5,v^5 tag{2}$$
Rozenbergs's lower bound when we substitute $x$ by $tsqrt(uv)$ 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.

(I consider it pertinent to add this answer to this challenging problem because the answer given here is different and simpler than the others given).

Lemma.-For every pair of positive $x,y$ it is verified that
$$frac{x^4}{8x^3+5y^3}+frac{y^4}{8y^3+5x^3}gefrac{x+y}{13}$$
Proof.- It is an easy algebraic operation leading to $$(x^3-y^3)(x^2-y^2)ge0$$
$$*******$$
Now, let $x,y,z$ be three arbitrary positive numbers. We have
$$frac{x^4}{8x^3+5y^3}+frac{y^4}{8y^3+5x^3}gefrac{x+y}{13}\frac{x^4}{8x^3+5z^3}+frac{z^4}{8z^3+5x^3}gefrac{x+z}{13}\frac{y^4}{8y^3+5z^3}+frac{z^4}{8z^3+5y^3}gefrac{y+z}{13}$$ It follows that the sum
$$left(frac{x^4}{8x^3+5y^3}+frac{y^4}{8y^3+5z^3}+frac{z^4}{8z^3+5x^3}right)+left(frac{x^4}{8x^3+5z^3}+frac{z^4}{8z^3+5y^3}+frac{y^4}{8y^3+5x^3}right)$$ is greater or equal than
$$frac{2(x+y+z)}{13}$$
These two expressions in parentheses can not be both smaller than $dfrac{x+y+z}{13}$ because their sum is greater than or equal to the double of this quantity. Thus one of them, say the first, satisfies
$$frac{x^4}{8x^3+5y^3}+frac{y^4}{8y^3+5z^3}+frac{z^4}{8z^3+5x^3}gefrac{x+y+z}{13}$$
Now the function $f:mathbb R^3_+rightarrow mathbb R_+$ defined by $f(x,y,z)=LHS$ is clearly continuous (it is even differentiable).

Choosing $(x,y,z)$ such that $0lt||(x,y,z)-(x,z,y)||to 0$ in $mathbb R^3_+$ we have that $|f(x,y,z)-f(x,z,y)|to 0$ in $mathbb R_+$. This ensures that if $f(x,y,z)gedfrac{x+y+z}{13}$ then this is so for $f(x,z,y)$ when $(x,z,y)$ is near enough of $(x,y,z)$. This ends the proof because from this we can deduce that one of the two expressions in parentheses above cannot be less to $dfrac{x+y+z}{13}$ if the other is greater or equal than $dfrac{x+y+z}{13}$ (which has been proved above).

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 $2times 3$ 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...

At the beginning we transform the original inequality:
$$dfrac{8x^4}{8x^3+5y^3}+dfrac{8y^4}{8y^3+5z^3}+dfrac{8z^4}{8z^3+5x^3}geq dfrac8{13}(x+y+z),$$
$$x-dfrac{5xy^3}{8x^3+5y^3}+y-dfrac{5yz^3}{8y^3+5z^3}+z-dfrac{5zx^3}{8z^3+5x^3}geq dfrac8{13}(x+y+z),$$
$$dfrac{xy^3}{8x^3+5y^3}+dfrac{yz^3}{8y^3+5z^3}+dfrac{zx^3}{8z^3+5x^3}leq dfrac1{13}(x+y+z).qquad(1)$$
Inequality $(1)$ is homogenius, and its decisions are determined up to an arbitrary common factor. We shall consider as the basic solution with
$$x+y+z=1.$$
It is easy to see that among the unknowns $x, y, z$ in $(1)$ only one can have zero limit.
Let $zto 0,$ then inequality transforms to
$$dfrac{xy^3}{8x^3+5y^3}leqdfrac1{13},quad x+y=1,$$
$$5y^3+8(1-y)^3-13y^3(1-y)geq 0.$$
Note that
$$5y^3+8(1-y)^3-13y^3(1-y) = 13y^4-16y^3+24 y^2-24y+8 \= 4y(1-y)^3 + 17y^4-28 y^3+36y^2-28y+8 = 4y(1-y)^3+7(1-y)^4 + 10y^4-6y^2+1 = 4y(1-y)^3+7(1-y)^4+(3y^2-1)^2+y^4 geq 0,$$
so the inequality is proved for this case .

For the case of
$$xnot=0,quad ynot=0,quad znot=0$$
we can use new method - of limiting transition.

I am not sure in its correctness, but discussion can be useful.

Let
$$f(t)=dfrac{t^3}{8+5t^3},$$
$$a = xfleft(dfrac{y}xright) + yfleft(dfrac{z}yright) + zfleft(dfrac{x}zright),$$
and find maximal possible value for $a.$

Using Cauchy-Schwarz inequality, we have
$$left(sqrt{x}cdotsqrt{x}fleft(dfrac{y}xright)+sqrt{y}cdotsqrt{y}fleft(dfrac{z}yright)+sqrt{z}cdotsqrt{z}fleft(dfrac{x}zright)right)^2 leq (x+y+z)left(xf^2left(dfrac{y}xright)+yf^2left(dfrac{z}yright)+zf^2left(dfrac{x}zright)right),$$
or
$$a^2 leq xf^2left(dfrac{y}xright)+yf^2left(dfrac{z}yright)+zf^2left(dfrac{x}zright),$$
and similarly
$$a^4 leq xf^4left(dfrac{y}xright) + yf^4left(dfrac{z}yright) + zf^4left(dfrac{x}zright),dots$$
$$a^{2^n} leq xf^{2^n}left(dfrac{y}xright) + yf^{2^n}left(dfrac{z}yright)+zf^{2^n}left(dfrac{x}zright)qquad(2)$$
for any $ninmathbb N.$

Let
$f_m = maxleft{fleft(dfrac{y}xright), fleft(dfrac{z}yright), fleft(dfrac{x}zright)right},$
then $(2)$ is equivalent to
$$left(dfrac{a}{f_m}right)^{2^n} leq xleft(dfrac{fleft(dfrac{y}xright)}{f_m}right)^{2^n} + yleft(dfrac{fleft(dfrac{z}yright)}{f_m}right)^{2^n} + zleft(dfrac{fleft(dfrac{x}zright)}{f_m}right)^{2^n}.qquad(3)$$

Inequality $(3)$ can be equality only if
$$lim_{ntoinfty}left(dfrac{a}{f_m}right)^{2^n} = lim_{ntoinfty}RHS.$$
If
$$fleft(dfrac{y}xright)not=fleft(dfrac{z}yright),quad fleft(dfrac{z}yright)not=fleft(dfrac{x}zright),quad fleft(dfrac{x}zright)not=fleft(dfrac{y}xright)$$
and for definiteness
$$f_m=fleft(dfrac{y}xright),$$
then for $a,x,y,z>0$ the system of conditions becomes inconsistent:
$$a=f_m,quad x=1,quad z=0.$$
If
$$f_m=fleft(dfrac{y}xright)=fleft(dfrac{z}yright)not=fleft(dfrac{x}zright),$$
or
$$f_m=fleft(dfrac{z}yright)=fleft(dfrac{x}zright)not=fleft(dfrac{y}xright),$$
or
$$f_m=fleft(dfrac{x}zright)=fleft(dfrac{y}xright)not=fleft(dfrac{z}yright),$$
one of the unknowns becomes zero, which reduces the problem to the case of the edges of area.

So remains the case of
$$f_m = fleft(dfrac{y}xright) = fleft(dfrac{z}yright) = fleft(dfrac{x}zright),$$
$$dfrac{y^3}{8x^3+5y^3}=dfrac{z^3}{8y^3+5z^3}=dfrac{x^3}{8z^3+5x^3},$$
or
$$5+8frac{x^3}{y^3} = 5+8frac{y^3}{z^3} = 5+8frac{z^3}{x^3},$$
$$dfrac xy = dfrac yz = dfrac zx, quad dfrac{y}xcdotdfrac{z}ycdotdfrac{x}z = 1, $$
$$x=y=z,$$
$$aleq f(1) = dfrac1{13},$$

$$boxed{dfrac{x^4}{8x^3+5y^3}+dfrac{y^4}{8y^3+5z^3}+dfrac{z^4}{8z^3+5x^3}geq dfrac{x+y+z}{13}}.$$

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 $dfrac{x}{z}={1over ab}$. We can suume without lose of generality that $a,ble1$ We need to prove that $$dfrac{x}{8+5left(dfrac{y}{x}right)^3}+dfrac{y}{8+5left(dfrac{z}{y}right)^3}+dfrac{z}{8+5left(dfrac{x}{z}right)^3}gedfrac{x+y+z}{13}$$by dividing the two sides of the inequality by $x$ and substituting $a,b,c$ we have that$$dfrac{1}{8+5left(dfrac{y}{x}right)^3}+dfrac{dfrac{y}{x}}{8+5left(dfrac{z}{y}right)^3}+dfrac{dfrac{z}{y}}{8+5left(dfrac{x}{z}right)^3}gedfrac{1+dfrac{y}{x}+dfrac{z}{x}}{13}$$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 $0le a,ble 1$

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.....

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 $nto3$ 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^2right)^{2}}{A}+frac{left(Y^2right)^{2}}{B}+frac{left(Z^2right)^{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^2right)^{2}}{A}+frac{left(Y^2right)^{2}}{B}+frac{left(Z^2right)^{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+Zright)}{(alpha+beta) left(X^{3} + Y^3+Z^3 right)} \
&=& frac{left(X+Y+Zright)}{(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 $beta=5$.

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) $$[8x^3+5y^3]+[8y^3+5z^3]+[8z^3+5x^3] = 13(x^3+y^3+z^3) geqslant 13(3 sqrt[3]{(xyz)^3}) = 13(3)$$

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, $x + y + z ge 3sqrt[3]{xyz} = 3$. So,
$$x^2 + y^2 + z^2 ge -3 + 3 + (x + y + z)$$
$$x^2 + y^2 + z^2 ge x + y + z ge 3$$

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 $sum_{cyc}frac{a^4}{8a^3+5b^3}geqsum_{cyc}frac{a^4}{8a^3+5c^3}$

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 $ageq 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 $sum_{cyc}frac{a^4}{8a^3+5b^3}leqsum_{cyc}frac{a^4}{8a^3+5c^3}$

The proof is the same as above .

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