# 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

• Cauchy-Schwarz via Titu may be an option... where's Michael Rozenberg when you need him? Commented May 14, 2020 at 16:09
• maybe use $\dfrac{x}{x+ 1} < 1$? Commented May 14, 2020 at 16:24
• @ab123,this equality $x=y=z=1$ is maximum of $\dfrac{1}{8}$,so if you use this $\dfrac{x}{x+1}<1$ is't right Commented May 14, 2020 at 16:27

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?

• Thank you,maybe there is C-S and AM-Gm inequality to solve it?because this is constant problem Commented May 15, 2020 at 0:29
• I knew you were hiding somewhere, Michael. Commented May 15, 2020 at 4:48

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 $$0 we have $$\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!