<s>[INST] Prove that $\sum\limits_{cyc}\frac{a}{a^{11}+1}\leq\frac{3}{2}$ for $a, b, c > 0$ with $abc = 1$ 
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.
 [/INST] This is probably wrong, but it might provide some ideas.
First observe that $a^nb^nc^n\leq3\;\;\forall n$, which is trivial by AM-GM.
Then, when $x\geq y$, then $\frac1x\leq\frac1y$. 
First, expand to get $$a(b^{11}+1)(c^{11}+1)+b(a^{11}+1)(c^{11}+1)+c(a^{11}+1)(b^{11}+1)\over(a^{11}+1)(b^{11}+1)(c^{11}+1)$$
so
$$a+b+c+ab(a^{10}+b^{10})+bc(b^{10}+c^{10})+ac(a^{10}+c^{10})+abc(a^{10}b^{10}+b^{10}c^{10}+a^{10}c^{10})\over(a^{11}+1)(b^{11}+1)(c^{11}+1)$$
By the first observation $a+b+c\geq3$. Also, by AM-GM, $a^{10}+b^{10}\geq\frac2{c^5}$. And since $ab=\frac1c$, we get
$$3+2\left(\frac1{a^6}+\frac1{b^6}+\frac1{c^6}\right)+(a^{10}b^{10}+b^{10}c^{10}+a^{10}c^{10})\over(a^{11}+1)(b^{11}+1)(c^{11}+1)$$
Then, by GM-HM, $$1\geq{3\over{(a^{10}b^{10}+b^{10}c^{10}+a^{10}c^{10})\over a^{10}b^{10}c^{10}}}$$ so $$(a^{10}b^{10}+b^{10}c^{10}+a^{10}c^{10})\geq3$$
Thus we get $$6+2\left(\frac1{a^6}+\frac1{b^6}+\frac1{c^6}\right)\over(a^{11}+1)(b^{11}+1)(c^{11}+1)$$
And similarly to above, with GM-HM we get $$\frac1{a^6}+\frac1{b^6}+\frac1{c^6}\geq3$$
Thus we get $$12\over(a^{11}+1)(b^{11}+1)(c^{11}+1)$$
And expanding the denominator we get $$12\over2+(a^{10}b^{10}+b^{10}c^{10}+a^{10}c^{10})+(a^{11}+b^{11}+c^{11})$$
And now I'm not sure what to do. Hope this helps.
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