Olympiad Inequality ∑cycx48x3+5y3⩾\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}

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

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”.

Answer

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

The Buffalo Way helps:

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

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

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

Done!

For example, we’ll prove that $$6(65u^3+189u^2v-176uv^2+65v^3)\geq531\sqrt{u^3v^3},6(65u^3+189u^2v-176uv^2+65v^3)\geq531\sqrt{u^3v^3},$$ which gives a coefficient $$531531$$ before $$t^7t^7$$ in the polynomial $$156t^8+531t^7+2t^6-632t^5-152t^4+867t^3+834t^2+299t+40.156t^8+531t^7+2t^6-632t^5-152t^4+867t^3+834t^2+299t+40.$$

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

Thus, we need to prove that:
$$130k^6+378k^4-177k^3-352k^2+130\geq0130k^6+378k^4-177k^3-352k^2+130\geq0$$
and by AM-GM we obtain: $$130k^6+378k^4-177k^3-352k^2+130=130k^6+378k^4-177k^3-352k^2+130=$$
$$=130\left(k^3+\frac{10}{13}k-1\right)^2+\frac{k}{13}(2314k^3+1079k^2-5576k+2600)\geq=130\left(k^3+\frac{10}{13}k-1\right)^2+\frac{k}{13}(2314k^3+1079k^2-5576k+2600)\geq$$
$$\geq\frac{k}{13}\left(8\cdot\frac{1157}{4}k^3+5\cdot\frac{1079}{5}k^2+21\cdot\frac{2600}{21}-5576k\right)\geq\geq\frac{k}{13}\left(8\cdot\frac{1157}{4}k^3+5\cdot\frac{1079}{5}k^2+21\cdot\frac{2600}{21}-5576k\right)\geq$$
$$\geq\frac{k^2}{13}\left(34\sqrt[34]{\left(\frac{1157}{4}\right)^8\left(\frac{1079}{5}\right)^5\left(\frac{2600}{21}\right)^{21}}-5576\right)>0.\geq\frac{k^2}{13}\left(34\sqrt[34]{\left(\frac{1157}{4}\right)^8\left(\frac{1079}{5}\right)^5\left(\frac{2600}{21}\right)^{21}}-5576\right)>0.$$
We’ll prove that $$2(377u^4+1206u^3v+585u^2v^2-1349uv^3+377v^4)\geq2u^2v^2, 2(377u^4+1206u^3v+585u^2v^2-1349uv^3+377v^4)\geq2u^2v^2,$$ for which it’s enough to prove that:
$$377t^4+1206t^3+584t^2-1349t+377\geq0377t^4+1206t^3+584t^2-1349t+377\geq0$$ or
$$t^4+\frac{1206}{377}t^3+\frac{584}{377}t^2-\frac{1349}{377}t+1\geq0t^4+\frac{1206}{377}t^3+\frac{584}{377}t^2-\frac{1349}{377}t+1\geq0$$ or
$$\left(t^2+\frac{603}{377}t-\frac{28}{29}\right)^2+\frac{131015t^2-69589t+9633}{142129}\geq0,\left(t^2+\frac{603}{377}t-\frac{28}{29}\right)^2+\frac{131015t^2-69589t+9633}{142129}\geq0,$$ which is true because
$$69589^2-4\cdot131015\cdot9633<0.69589^2-4\cdot131015\cdot9633<0.$$

Attribution
Source : Link , Question Author : HN_NH , Answer Author : Michael Rozenberg