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.