TITLE: Prove that: $2(ab+bc+ca)-a^2-b^2-c^2\le6$.
QUESTION [5 upvotes]: 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: $2(ab+bc+ca)-a^2-b^2-c^2\le6$.

I have no idea for solve this problem.

REPLY [2 votes]: $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 $