prove that : $\frac{a^2+b^2}{a+b} + \frac{b^2+c^2}{b+c}+ \frac{c^2+a^2}{c+a} \geq 3$ 
For $a^2+b^2+c^2 =3$, with $a$, $b$ and $c$ are positive real numbers, prove that:
  $$\frac{a^2+b^2}{a+b} + \frac{b^2+c^2}{b+c}+ \frac{c^2+a^2}{c+a} \geq 3.$$

Can any one help me with this problem?

Multiplying both LHS and RHS by $a+b+c = s > 0$, you have:
$$\sum_{cyc}\left(\frac{a^2+b^2}{a+b}\cdot (a+b) \right)+ \sum_{cyc}\left(\frac{a^2+b^2}{a+b}\cdot c \right) \ge 3(a+b+c) $$
$$\iff 6+ \sum_{cyc}\left(\frac{a^2+b^2}{a+b}\cdot c \right) \ge 3s $$
Now note that $a^2+b^2 \ge \frac12(a+b)^2$ using AM-GM, so it is enough to show that
$$6 + \frac12\sum_{cyc}(s-c)c \ge 3s \iff 6 + \frac{s^2}2-\frac32 \ge 3s \iff (s-3)^2\ge 0$$
