Find the Maximum

The following solution isn't mine.

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The condition is equivalent to $ab+bc+ca= \dfrac{1}{3}$ .

I claim that the expression never exceeds $3$ . This is equivalent to showing that $(ab+bc+ca )\left(\dfrac{1}{a^2+1-bc} + \dfrac{1}{b^2+1-ca} + \dfrac{1}{c^2+1-ab} \right) \leq 1 \\ \iff \displaystyle \sum_{\text{cyc}} \dfrac{2(ab+bc+ca)}{a^2-bc+1} \leq 2 \\ \iff 3 - \displaystyle \sum_{\text{cyc}} \dfrac{2ab+2bc+2ca}{a^2-bc+1} \geq 1 \\ \iff \displaystyle \sum_{\text{cyc}} 1 - \dfrac{2(ab+bc+ca)}{a^2-bc+1} \geq 1 \\ \iff \displaystyle \sum_{\text{cyc}} \dfrac{a^2(a+b+c)}{a^3-abc+a} \geq 1.$ By Titu's lemma, $\displaystyle \sum_{\text{cyc}} \dfrac{a^2}{a^3-abc+a} \geq \dfrac{(a+b+c)^2}{a^3+b^3+c^3 - 3abc + a+b+c}.$ Hence, it suffices to show that $(a+b+c)^3 \geq a^3+b^3+c^3-3abc+a+b+c ,$ which is, in fact, an identity. Hence, we're done.

Equality holds for $a=b=c=\dfrac{1}{\sqrt{3}}.$

Hence, our desired answer is $\boxed{3}$ .

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This problem is from a recent China TST.

There are many elegant ways to solve this, but I am not at all in the moooooooooooood .

$\color{#D61F06}{\text{Equality case gives the extreme values!}}$

$a=b=c$ gives each of the fractions to be $1$ , giving answer $3$ .