If $a,b,c >0$ and $$ \frac{1}{a^{2} + b^{2} + 1} + \frac{1}{b^{2} + c^{2} + 1} + \frac{1}{c^{2} + a^{2} + 1} \ge 1 $$ prove that $ab + ac + bc \le 3$
Attempt:
We can show that $\frac{1}{3} \le \frac{1}{ab+bc+ac}$. Or we can show that: $a^{2} + b^{2} + c^{2} \le 3$ (Since by $AM-GM$, $ab + ac + bc \le a^{2} + b^{2} + c^{2}$).
Notice that $$ \frac{1}{a^{2} + b^{2} + 1} + \frac{1}{b^{2} + c^{2} + 1} + \frac{1}{c^{2} + a^{2} + 1} \ge 1 $$
implies
$$ 2 \ge \frac{a^{2} + b^{2}}{a^{2} + b^{2} + 1} + \frac{b^{2} + c^{2}}{b^{2} + c^{2} + 1} + \frac{a^{2} + c^{2}}{c^{2} + a^{2} + 1} $$
By Cauchy-Schwartz inequality: $ (a^2+b^2+1)(1+1+c^2) \ge (a+b+c)^2$, which impies that $1\le \sum \frac{1}{a^2+b^2+1}=\sum\frac{2+c^2}{(a^2+b^2+1)(1+1+c^2)}\le\sum\frac{c^2+2}{(a+b+c)^2}=\frac{a^2+b^2+c^2+6}{(a+b+c)^2}\implies(a+b+c)^2\le a^2+b^2+c^2+6\implies a^2+b^2+c^2+2(ab+bc+ca)\le a^2+b^2+c^2+6 \implies ab+bc+ca\le3$