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

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  • $\begingroup$ Oh, I feel that inequalities whose datums have $abc=1$ are more difficult than others. $\endgroup$ – Tĩnh Thu Jan 2 '14 at 7:03
  • $\begingroup$ maybe you forget to write a plus sign? $\endgroup$ – mathlove Jan 2 '14 at 7:06
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    $\begingroup$ Oh, I forgot. Thanks. $\endgroup$ – Tĩnh Thu Jan 2 '14 at 7:09
  • $\begingroup$ Out of curiosity, what math class has this type of problem as homework? $\endgroup$ – DanielV Jan 2 '14 at 9:33
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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 $1$.

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  • $\begingroup$ it is only the guess, not proof. $\endgroup$ – chenbai Jan 3 '14 at 5:08
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the Max should be $1$, so we need to prove:

$\dfrac{a}{b^2+c^2+a}+\dfrac{b}{c^2+a^2+b}+\dfrac{c}{a^2+b^2+c} \le 1 \iff \dfrac{b^2+c^2}{b^2+c^2+a}+\dfrac{c^2+a^2}{c^2+a^2+b}+\dfrac{a^2+b^2}{a^2+b^2+c} \ge 2$

$\dfrac{b^2+c^2}{b^2+c^2+a} \ge \dfrac{2bc}{2bc+a} =\dfrac{2}{2+a^2} \implies \sum \dfrac{1}{2+a^2} \ge 1 \iff a^2b^2+b^2c^2+c^2a^2 \ge3$

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  • $\begingroup$ $ \sum \dfrac{1}{2+a^2} \ge 1 $ is not true, instead $ \sum \dfrac{1}{2+a^2} \le 1 $ is true. They are equivalent to $a^2b^2+b^2c^2+c^2a^2 \le3$ and $a^2b^2+b^2c^2+c^2a^2 \ge3$ respectively. $\endgroup$ – didgogns Feb 7 '19 at 13:55
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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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