If $a$, $b$ and $c$ are three positive real numbers, prove that $\frac{a^2+1}{b+c}+\frac{b^2+1}{c+a}+\frac{c^2+1}{a+b} ≥ 3$ If $a$, $b$ and $c$ are three positive real numbers, prove that $$\frac{a^2+1}{b+c}+\frac{b^2+1}{c+a}+\frac{c^2+1}{a+b} ≥ 3$$
I think we are supposed to use the AM-GM inequality here; or maybe this could even be solved using the $T_2$'s Lemma, although I'm a bit skeptical about that. 

After using AM-GM we need to prove that
$$(a^2+1)(b^2+1)(c^2+1)\geq(a+b)(a+c)(b+c)$$
which follows from C-S: $(a^2+1)(1+b^2)\geq(a+b)^2$.
Done!
