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
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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!
Now, let $x=\frac{a}{b},$ $y=\frac{b}{c}$, where $a$, $b$ and $c$ are positives.
Thus, $z=\frac{c}{a}$ and since $$x^x\geq x,$$ it's enough to prove that: $$\sum_{cyc}\frac{ab}{13a^2+10ab+b^2}\leq\frac{1}{8},$$ which is true by BW.
Indeed, let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.
Thus, we need to prove that: $$384(u^2-uv+v^2)a^4+192(2u^3+7u^2v-5uv^2+2v^3)a^3+$$ $$+16(13u^4+82u^3v+39u^2v^2-62uv^3+13v^4)a^2+$$ $$+4(91u^3+258u^2v-162uv^2+13v^3)uva+13(13u^2+2uv+v^2)u^2v^2\geq0.$$ Now, we can show that: $$384(u^2-uv+v^2)\geq384uv,$$ $$192(2u^3+7u^2v-5uv^2+2v^3)\geq768\sqrt{u^3v^3},$$ $$16(13u^4+82u^3v+39u^2v^2-62uv^3+13v^4)\geq-32u^2v^2,$$ $$4(91u^3+258u^2v-162uv^2+13v^3)\geq-384\sqrt{u^3v^3}$$ and $$13(13u^2+2uv+v^2)\geq112uv.$$ Now, let $a=t\sqrt{uv}.$
Thus, it's enough to prove that: $$384t^4+768t^3-32t^2-384t+112\geq0,$$ which is smooth.
Can you end it now?