Prove that: $2(ab+bc+ca)-a^2-b^2-c^2\le6$.

Question

Let $a,b,c>0$ such that: $\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}=1$.

Prove that: $2(ab+bc+ca)-a^2-b^2-c^2\le6$.

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I have no idea for solve this problem.

Answer 1

$2=\dfrac{a^2}{1+a^2}+\dfrac{b^2}{1+b^2}+\dfrac{c^2}{1+c^2}\ge \dfrac{(a+b+c)^2}{a^2+b^2+c^2+3} \iff 2(a^2+b^2+c^2+3)\ge (a+b+c)^2 \iff 2(ab+bc+ac)-a^2-b^2-c^2\le6 $