TITLE: For $abc=1$ prove that $\sum\limits_{cyc}\frac{a}{a^{11}+1}\leq\frac{3}{2}.$
QUESTION [17 upvotes]: Let $a$, $b$ and $c$ be positive numbers such  that $abc=1$. Prove that:
$$\frac{a}{a^{11}+1}+\frac{b}{b^{11}+1}+\frac{c}{c^{11}+1}\leq\frac{3}{2}.$$
I tried homogenization and the BW (https://artofproblemsolving.com/community/c6h522084),
but it does not work.
Indeed, let $a=\frac{x}{y}$, $b=\frac{y}{z}$, where $x$, $y$ and $z$ are positives.
Hence, $c=\frac{z}{x}$ and we need to prove that $$\sum_{cyc}\frac{xy^{10}}{x^{11}+y^{11}}\leq\frac{3}{2},$$
which has a problem around $(x,y,z)=(7,5,6)$.
For these values $$\frac{3}{2}-\sum_{cyc}\frac{xy^{10}}{x^{11}+y^{11}}=0.0075...$$
I tried also TL, uvw, C-S, Lagrange multipliers and more, but without success. 
Also, Vasc's Theorems don't help.
Also, the following method does not help here. Find the maximum of the expression
Because the inequality $\frac{x}{x^{11}+1}\leq\frac{3(a^9+1)}{4(a^{18}+a^9+1)}$ is wrong.

REPLY [4 votes]: We start with the case $a\leq 1$ , $b\leq 1$ , $c\geq 1$ so we have to prove this :
$$\frac{a}{a^{11}+1}+\frac{b}{b^{11}+1}+\frac{c}{c^{11}+1}\leq1.5$$
Or with $a\geq 1$, $b\geq 1$ , $c\leq 1$ :
$$\frac{a^{10}}{a^{11}+1}+\frac{b^{10}}{b^{11}+1}+\frac{c^{10}}{c^{11}+1}\leq1.5$$
We have the following lemma :

Let $a,b$ be real positive numbers with $a\geq 1$, $b\geq 1$ then we have :
  $$\frac{a^{10}}{a^{11}+1}+\frac{b^{10}}{b^{11}+1}\leq \frac{a+b}{ab}\frac{(\frac{2ab}{a+b})^{11}}{(\frac{2ab}{a+b})^{11}+1}$$

Proof :
It's just the inequality of Jensen apply to $f(x)$ wich is concave for $x\geq 1$ :
$f(x)=\frac{x^{11}}{x^{11}+1}$
With coefficient :
$\alpha_1=\frac{1}{a}\frac{ab}{a+b}$
And 
$\alpha_2=\frac{1}{b}\frac{ab}{a+b}$
So we have to prove :
$$\frac{c^{10}}{c^{11}+1}+\frac{a+b}{ab}\frac{(\frac{2ab}{a+b})^{11}}{(\frac{2ab}{a+b})^{11}+1}\leq 1.5$$
We have this other lemma :

$$\frac{c^{10}}{c^{11}+1}=\frac{ab}{(ab)^{11}+1}\leq \frac{(\frac{2ab}{a+b})^{2}}{(\frac{2ab}{a+b})^{22}+1} $$

Proof :
It's easy to show this because $f(x)=\frac{x}{x^{11}+1}$ is decreasing for $x\geq 1$
It's remains to prove :
$$(\frac{2ab}{a+b})^{2}\leq ab $$
Or :
$$ab\leq 0.5(a+b)^2 $$
Wich is obvious.
So we have to prove this :
$$\frac{(\frac{2ab}{a+b})^{2}}{(\frac{2ab}{a+b})^{22}+1}+\frac{a+b}{ab}\frac{(\frac{2ab}{a+b})^{11}}{(\frac{2ab}{a+b})^{11}+1}\leq 1.5$$
But we put :
$x=\frac{2ab}{a+b}$
We get :
$$\frac{2x^{10}}{x^{11}+1}+\frac{x^2}{x^{22}+1}\leq 1.5$$
Wich is true !
The other case wich is $a\leq 1$ , $b\geq 1$ , $c\geq 1$ and :
$$\frac{a}{a^{11}+1}+\frac{b}{b^{11}+1}+\frac{c}{c^{11}+1}\leq1.5$$
Is easy to show because :
$$\frac{a}{a^{11}+1}+\frac{b}{b^{11}+1}+\frac{c}{c^{11}+1}\leq\frac{a}{a^{11}+1}+\frac{b^{10}}{b^{11}+1}+\frac{c}{c^{11}+1}$$
Done !
Edit :
We prove that for $x\geq 1$:
$$\frac{2x^{10}}{x^{11}+1}+\frac{x^2}{x^{22}+1}\leq 1.5$$
First we prove that :
$$\frac{2x^{10}}{x^{11}+1}+\frac{x^2}{x^{22}+1}\leq 2$$
We have the following identity :
$$\frac{2}{x}-\frac{2}{x(x^{11}+1)}+\frac{1}{x^{20}}-\frac{1}{x^{20}(x^{22}+1)}=\frac{2x^{10}}{x^{11}+1}+\frac{x^2}{x^{22}+1}$$
So we have to prove this :
$$\frac{2}{x}+\frac{1}{x^{20}}\leq\frac{2}{x(x^{11}+1)}+\frac{1}{x^{20}(x^{22}+1)}+2$$
Wich is obvious because we have :
$$\frac{2}{x}\leq 2$$
And 
$$\frac{1}{x^{20}}\leq \frac{2}{x(x^{11}+1)}$$
So we have proved :
$$\frac{2x^{10}}{x^{11}+1}+\frac{x^2}{x^{22}+1}\leq 2$$
Or 
$$\frac{x^{10}}{x^{11}+1}+\frac{0.5x^2}{x^{22}+1}\leq 1$$
Now we put :
$f(x)=\frac{x^{10}}{x^{11}+1}+\frac{0.5x^2}{x^{22}+1}$
$g(x)=\frac{x^{10}}{x^{11}+1}$
$h(x)=\frac{0.5x^2}{x^{22}+1}$
And we want to prove this :
$$\frac{f(x)^{11}}{f(x)^{11}+1}\leq \frac{0.75^{11}}{0.75^{11}+1}$$
First of all the function $u(x)=\frac{x^{11}}{x^{11}+1}$ is convex for $x\leq 1$ so we put $x=\frac{1}{y}$ and :
$g(x)=\frac{x^{10}}{x^{11}+1}=\frac{y}{y^{11}+1}=l(y)$
$h(x)=\frac{0.5x^2}{x^{22}+1}=\frac{0.5y^{20}}{y^{22}+1}=p(y)$
$$f(x)=q(y)=\frac{y}{y^{11}+1}+\frac{0.5y^{20}}{y^{22}+1}$$
It remains to prove :
$$\frac{q(y)^{11}}{q(y)^{11}+1}\leq \frac{0.75^{11}}{0.75^{11}+1}$$
We can apply Jensen's inequality to $u(x)$ (because all the value are inferior to one) to get :
$$\frac{q(y)^{11}}{q(y)^{11}+1}\leq [\frac{\alpha}{\alpha+\beta}\frac{(\frac{l(y)}{\alpha})^{11}}{(\frac{l(y)}{\alpha})^{11}+1}+\frac{\beta}{\alpha+\beta}\frac{(\frac{p(y)}{\beta})^{11}}{(\frac{p(y)}{\beta})^{11}+1}]$$
We put finally :
$$\alpha=\frac{4l(y)}{3}$$
And 
$$\beta=\frac{4p(y)}{3}$$
To get :
$$\frac{q(y)^{11}}{q(y)^{11}+1}\leq \frac{0.75^{11}}{0.75^{11}+1}$$
But the function $u(x)$ is increasing so we have :
$$q(y)\leq \frac{3}{4}$$
Or 
$$\frac{2x^{10}}{x^{11}+1}+\frac{x^2}{x^{22}+1}\leq 1.5$$
Done !