<s>[INST] 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
$$M=\frac{a}{b^2+c^2+a}+\frac{b}{c^2+a^2+b}+\frac{c}{a^2+b^2+c}.$$
 [/INST] 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!
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