An olympiad-like inequality $\frac{x^5}{6x^4+5y^4}+\frac{y^5}{6y^4+5z^4}+\frac{z^5}{6z^4+5x^4}\geq \frac{x+y+z}{11}$

buffalo-wayinequalityjensen-inequality

Hi it's an inequality in the spirit of the Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$ :

Let $x,y,z>0$ then we have :
$$\frac{x^5}{6x^4+5y^4}+\frac{y^5}{6y^4+5z^4}+\frac{z^5}{6z^4+5x^4}\geq \frac{x+y+z}{11}$$

My partial answer :

Since we can rewrite the inequality as :
$$\frac{x}{x+y+z}\frac{1}{6+5\frac{y^4}{x^4}}+\frac{y}{x+y+z}\frac{1}{6+5\frac{z^4}{y^4}}+\frac{z}{x+y+z}\frac{1}{6+5\frac{x^4}{z^4}}\geq \frac{1}{11}$$

We can apply Jensen's inequality with $f(x)=\frac{1}{6+5x^4}$ on $[ 2^{0.25}\sqrt{\frac{3}{5}},\infty[ $

And it works for $\frac{y}{x},\frac{z}{y},\frac{x}{z}\in[ 2^{0.25}\sqrt{\frac{3}{5}},\infty[ $

I have tried also derivatives and I can reduce this three variables inequality to a one variable inequality but we get a little monster at the end .

I'm not familiar with the "Buffalo's way" and maybe it works with that .

So thanks a lot for sharing answers and comments .

Best Answer

Yes, the BW helps!

For $x\geq y\geq z$ we have $$\sum_{cyc}\left(\frac{x^5}{6x^4+5y^4}-\frac{x^5}{6x^4+5z^4}\right)=(x-y)(x-z)(y-z)P(x,y,z)\geq0$$ because $P(x,y,z)>0.$ See here:

https://www.wolframalpha.com/input/?i=x%5E5%2F%286x%5E4%2B5y%5E4%29%2By%5E5%2F%286y%5E4%2B5z%5E4%29%2Bz%5E5%2F%286z%5E4%2B5x%5E4%29-%28x%5E5%2F%286x%5E4%2B5z%5E4%29%2By%5E5%2F%286y%5E4%2B5x%5E4%29%2Bz%5E5%2F%286z%5E4%2B5y%5E4%29%29

Thus, it's enough to prove our inequality for $x\leq y\leq z$.

Now, let $y=x+u$ and $z=x+u+v$.

Thus, we need to prove that: $$308(u^2+uv+v^2)x^{11}+2(1155u^3+783u^2v+706uv^2+539v^3)x^{10}+$$ $$+4(2134u^4+1103u^3v-78u^2v^2+953uv^3+594v^4)x^9+$$ $$+(20449u^5+12709u^4v-11300u^3v^2-1174u^2v^3+8876uv^4+3245v^5)x^8+$$ $$+2(17314u^6+16686u^5v-10645u^4v^2-11008u^3v^3+5199u^2v^4+6190uv^5+1386v^6)x^7+$$ $$+2(21297u^7+30814u^6v-2994u^5v^2-17467u^4v^3+2704u^3v^4+11205u^2v^5+4998uv^6+726v^7)x^6+$$ $$+(38225u^8+75932u^7v+30362u^6v^2-12472u^5v^3+14201u^4v^4+33980u^3v^5+19236u^2v^6+4584uv^7+429v^8)x^5+$$ $$+(24816u^9+62916u^8v+51728u^7v^2+23842u^6v^3+36690u^5v^4+47975u^4v^5+28812u^3v^6+8292u^2v^7+1089uv^8+55v^9)x^4+$$ $$+2u(5687u^9+17490u^8v+20790u^7v^2_17568u^6v^3+21756u^5v^4+24360u^4v^5+15750u^3v^6+5400u^2v^7+885uv^8+50v^9)x^3+$$ $$+6u^2(u+v)^2(583u^7+935u^6v+737u^5v^2+1155u^4v^3+1485u^3v^4+915u^2v^5+255uv^6+25v^7)x^2+$$ $$+u^3(u+v)^3(649u^6+729u^5v+750u^4v^2+1490u^3v^3+1485u^2v^4+645uv^5+100v^6)x+$$ $$+5u^4(u+v)^4(11u^5+7u^4v+14u^3v^32+26u^2v^3+19uv^4+5v^5)\geq0,$$ which is true because all polynomials of $u$ and $v$ before powers of $x$ are non-negative.

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