<s>[INST] Inequality with $xyz=1$ Given $x,y,z>0$ such that $xyz=1$
Prove that: $\frac{x^7}{x^8+1}+\frac{y^7}{y^8+1}+\frac{z^7}{z^8+1}\leq \frac{3}{2}$
P/s: I tried to solve it by AM-HM but I failed
 [/INST] since
\begin{align*}
&\frac34-\frac34\cdot\frac1{x^{12}+x^6+1}-\frac{x^7}{x^8+1}\\
={}&\frac{x^6(x-1)^2(3x^{12}+2x^{11}+x^{10}-x^8-2x^7-2x^5-x^4+x^2+2x+3)}{4(x^{12}+x^6+1)(x^8+1)}
\end{align*}
and
\begin{align*}
2x^{11}+2x&\geqslant2x^7+2x^5,\\
x^{10}+x^2&\geqslant x^8+x^4,
\end{align*}
we get
$$\frac{x^7}{x^8+1}\leqslant\frac34-\frac34\cdot\frac1{x^{12}+x^6+1},$$
so, just need to prove
$$\frac94-\frac34\sum\frac1{x^{12}+x^6+1}\leqslant\frac32,$$
or
$$\sum\frac1{x^{12}+x^6+1}\geqslant1,$$
or equivalent to: $a$, $b$, $c>0$, $abc=1$, prove that
$$\sum\frac1{a^2+a+1}\geqslant1,$$
which is well-known.
</s>