Find max: $M=\frac{a}{b^2+c^2+a}+\frac{b}{c^2+a^2+b}+\frac{c}{a^2+b^2+c}$ For $a,b,c>0$ and $abc=1$, find the maximum of
$$M=\frac{a}{b^2+c^2+a}+\frac{b}{c^2+a^2+b}+\frac{c}{a^2+b^2+c}.$$

Remark: Perhaps there are alternative proofs of $(ab)^{3/2} + (bc)^{3/2} + (ca)^{3/2} \ge ab + bc + ca$ without using Chebyshev sum inequality.
By Cauchy-Bunyakovsky-Schwarz inequality, we have
$$M = \sum_{\mathrm{cyc}} \frac{a(b + c + a^2)}{(b^2 + c^2 + a)(b + c + a^2)}
\le \sum_{\mathrm{cyc}} \frac{a(b + c + a^2)}{(a^{3/2} + b^{3/2} + c^{3/2})^2}$$
$$= \frac{a^3 + b^3 + c^3 + 2ab + 2bc + 2ca}{(a^{3/2} + b^{3/2} + c^{3/2})^2}$$
and
$$1 - M \ge \frac{2ab(\sqrt{ab} - 1) + 2bc(\sqrt{bc} - 1) + 2ca(\sqrt{ca} - 1)}{(a^{3/2} + b^{3/2} + c^{3/2})^2} \ge 0$$
where we have used Chebyshev sum inequality to get
\begin{align*}
 &2ab(\sqrt{ab} - 1) + 2bc(\sqrt{bc} - 1) + 2ca(\sqrt{ca} - 1)\\
 \ge\,& \frac13(2ab + 2bc + 2ca)(\sqrt{ab} + \sqrt{bc} + \sqrt{ca} - 3)\\
 \ge\,& 0.
\end{align*}
(Note: $2ab, 2bc, 2ca$ and $\sqrt{ab}-1, \sqrt{bc}-1, \sqrt{ca} - 1$ are in a similar order.)
Also, when $a = b = c = 1$, we have $M = 1$.
Thus, the maximum of $M$ is $1$.
