Show that $\frac{a^2+1}{b+c}+\frac{b^2+1}{a+c}+\frac{c^2+1}{a+b} \ge 3$ 
If $a,b,c$ are positive numbers then show that 
  $$\dfrac{a^2+1}{b+c}+\dfrac{b^2+1}{a+c}+\dfrac{c^2+1}{a+b} \ge 3$$

I am stuck at the first stage. Please give me some hints so that I can solve the problem. Thanks in advance. 

Hint: Note that $a^2+1 \ge 2a$ and also use NESBIT Inequality.
Note that  $(a-1)^2\ge 0 \implies a^2+1 \ge 2a$
Otherwise use $AM \ge GM $ for $a^2$ and $1$, you get $a^2+1 \ge 2a$.
NESBITT Inequality: $$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \ge \dfrac 3 2$$
For proof of NESBITT Inequality you can see here.
