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. 

Let us define $t=a+b+c$. First we note that
$$\frac{a^2+1}{b+c}+\frac{b^2+1}{c+a}+\frac{c^2+1}{a+b}=
\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)+
\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)$$
so by Titu's lemma:
$$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\geq\frac{(a+b+c)^2}{2(a+b+c)}=\frac{t}{2}$$
and 
$$\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\geq\frac{3^2}{2(a+b+c)}=\frac{9}{2t}.$$
Therefore, we have to check that 
$$\frac{t}{2}+\frac{9}{2t}\geq3$$
for $t>0$. But this follows from the fact that the polynomial $p(t)=t^2-6t+9=(t-3)^2\geq0$.
