Is there a direct proof of the inequality $cyclic \sum\frac {a^2+1}{b+c}\ge 3$ Let $a,b,c>0$ reals.
Prove that
$$cyclic \sum\frac {a^2+1}{b+c}\ge 3$$
I proved it using Nesbitt inequality
$$cyclic \sum \frac {a}{b+c} \ge \frac {3}{2} $$
and the fact that
$$a+\frac {1}{a}\ge 2$$
But i would like to know if there is a straight proof without Nesbitt inequality.

By AM-GM and C-S we obtain:
$$\sum_{cyc}\frac{a^2+1}{b+c}=\sum_{cyc}\frac{a^2}{b+c}+\sum_{cyc}\frac{1}{b+c}\geq2\sqrt{\sum_{cyc}\frac{a^2}{b+c}\sum_{cyc}\frac{1}{b+c}}\geq$$
$$\geq2\sqrt{\frac{(a+b+c)^2}{\sum\limits_{cyc}(b+c)}\sum_{cyc}\frac{1}{b+c}}=\sqrt{2(a+b+c)\sum_{cyc}\frac{1}{b+c}}=$$
$$=\sqrt{\sum\limits_{cyc}(b+c)\sum_{cyc}\frac{1}{b+c}}\geq\sqrt{(1+1+1)^2}=3.$$
