<s>[INST] Prove this inequality with $xyz=1$ let $x,y,z>0$ and such $xyz=1$,show that
$$f(x)+f(y)+f(z)\le\dfrac{1}{8}$$
where $f(x)=\dfrac{x}{2x^{x+1}+11x^2+10x+1}$
I try use this $2x^x\ge x^2+1$,so we have
$$2x^{x+1}+11x^2+10x+1\ge x^3+11x^2+11x+1=(x+1)(x^2+10+1)$$
It need to prove
$$\sum_{cyc}\dfrac{x}{(x+1)(x^2+10x+1)}\le\dfrac{1}{8},$$where $xyz=1$
then I can't 
 [/INST] The TL method helps! 
Since $$x^x\geq\frac{x^3-x^2+x+1}{2},$$ it's enough to prove that
$$\sum_{cyc}\frac{x}{x^4-x^3+12x^2+11x+1}\leq\frac{1}{8}$$ or
$$\sum_{cyc}\left(\frac{1}{24}-\frac{x}{x^4-x^3+12x^2+11x+1}\right)\geq0$$ or
$$\sum_{cyc}\left(\frac{1}{24}-\frac{x}{x^4-x^3+12x^2+11x+1}-\frac{1}{48}\ln{x}\right)\geq0.$$
Now, prove that for any $0<x<6$ we have $$\frac{1}{24}-\frac{x}{x^4-x^3+12x^2+11x+1}-\frac{1}{48}\ln{x}\geq0,$$
which says that our inequality is proven for $\max\{x,y,z\}<6.$
Let $x\geq6$.
Now, we see that for any $x>0$ we have $$\frac{x}{x^4-x^3+12x^2+11x+1}\leq\frac{1}{17}$$ and for any $x\geq6$ we have $$\frac{x}{x^4-x^3+12x^2+11x+1}\leq\frac{6}{6^4-6^3+12\cdot6^2+11\cdot6+1}\leq\frac{6}{1579}.$$
Id est,
$$\sum_{cyc}\frac{x}{x^4-x^3+12x^2+11x+1}\leq\frac{6}{1579}+\frac{2}{17}<\frac{1}{8}$$ and we are done!
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