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
$$M=\frac{a}{b^2+c^2+a}+\frac{b}{c^2+a^2+b}+\frac{c}{a^2+b^2+c}.$$
 A: the Max should be $1$, so we need to prove:
$\dfrac{a}{b^2+c^2+a}+\dfrac{b}{c^2+a^2+b}+\dfrac{c}{a^2+b^2+c} \le 1 \iff \dfrac{b^2+c^2}{b^2+c^2+a}+\dfrac{c^2+a^2}{c^2+a^2+b}+\dfrac{a^2+b^2}{a^2+b^2+c} \ge 2$
$\dfrac{b^2+c^2}{b^2+c^2+a} \ge \dfrac{2bc}{2bc+a} =\dfrac{2}{2+a^2} \implies \sum \dfrac{1}{2+a^2} \ge 1 \iff a^2b^2+b^2c^2+c^2a^2 \ge3$
A: Take the first term:$$\frac{a}{b^2+c^2+a}=\frac{1}{\frac{1}{a^2}(\frac{b}{c}+\frac{c}{b})+1}$$
Minimum value of $\frac{1}{a^2}(\frac{b}{c}+\frac{c}{b})$ is $\frac{2}{a^2}$ which it takes when $b=c$. By the symmetry of the expression it is maximum when $a=b=c$ and the maximum value is $1$.
A: Substitute $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{x}$ for positive numbers $x,y,z$. We will show $M\le1$, which is equivalent to$$\sum_{cyc}\frac{x^3z^2}{x^2y^3+yz^4+x^3z^2}\le1$$and after full expansion, it is equivalent to$$\sum_{cyc}x^8 y^3 z^4 + x^7 y^2 z^6\le\sum_{cyc}x^9y^3z^3+x^7y^7z$$
By rearrangement inequality, it is obvious that $\sum_{cyc}x^5z\le\sum_{cyc}x^6$ hence $\sum_{cyc}x^8y^3z^4\le\sum_{cyc}x^9y^3z^3$. Also, for variables $p,q,r$, $\sum_{cyc}p^5r\le\sum_{cyc}p^6$ is true. Let $p=xy$, $q=xz$, $r=yz$ then it is $\sum_{cyc}x^5y^6z\le\sum_{cyc}x^6y^6$. Therefore $\sum_{cyc}x^7y^2z^6=\sum_{cyc}x^6y^7z^2\le\sum_{cyc}x^7y^7z$ and it is proved!
A: 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 $1$.
