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. 
 A: 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.
A: 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.
A: Setting $a=b=c=x$ and then using $ x^2+1 \ge x $ gives the required result.
This seems true for a lot of symmetric inequalities. I wonder if there is a Grand Theorem for Symmetric Inequalities?
A: By AM-GM and C-S we obtain:
$$\sum_{cyc}\frac{a^2+1}{b+c}\geq3\sqrt[3]{\frac{\prod\limits_{cyc}(a^2+1)}{\prod\limits_{cyc}(a+b)}}=3\sqrt[3]{\frac{\sqrt{\prod\limits_{cyc}\left((a^2+1)(1+b^2)\right)}}{\prod\limits_{cyc}(a+b)}}\geq$$
$$\geq3\sqrt[3]{\frac{\sqrt{\prod\limits_{cyc}\left(a+b\right)^2}}{\prod\limits_{cyc}(a+b)}}=3$$
