<s>[INST] 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. 
 [/INST] Direct Question of Titu's Lemma.
$\dfrac{1}{b+c}+ \dfrac{1}{c+a}+\dfrac{1}{a+b} \ge \dfrac{(1+1+1)^2}{2(a+b+c)}$
Similarly,
$\dfrac{a^2}{b+c}+ \dfrac{b^2}{c+a}+\dfrac{c^2}{a+b} \ge \dfrac{(a+b+c)^2}{2(a+b+c)}$
Let, $a+b+c=k$
$\dfrac{3^2}{2k}+\dfrac{k}{2} \ge 2\sqrt{3^2 \cdot \dfrac{k}{2} \dfrac{1}{2k}} = 3$, we are done.
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