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?

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that
$$\sum_{cyc}\frac{a^2+b^2}{a+b}\geq\sqrt{3(a^2+b^2+c^2)}$$ or
$$\frac{\sum\limits_{cyc}(a^2+b^2)(c^2+ab+ac+bc)}{\prod\limits_{cyc}(a+b)}\geq3\sqrt{3u^2-2v^2}$$ or
$$\frac{\sum\limits_{cyc}(a^2+b^2)(c^2+3v^2)}{9uv^2-w^3}\geq3\sqrt{3u^2-2v^2}$$ or
$$\frac{2(9v^4-6uw^3)+6v^2(9u^2-6v^2)}{9uv^2-w^3}\geq3\sqrt{3u^2-2v^2}$$ or
$$2(9u^2v^2-3v^4-2uw^3)\geq(9uv^2-w^3)\sqrt{3u^2-2v^2}$$ or
$$ 2(9u^2v^2-3v^4)-9uv^2\sqrt{3u^2-2v^2}\geq(4u-\sqrt{3u^2-2v^2})w^3$$
and since $4u-\sqrt{3u^2-2v^2}>0$ it's enough to prove our inequality 
for a maximal value of $w^3$, which happens for equality case of two variables.
Since $\sum\limits_{cyc}\frac{a^2+b^2}{a+b}\geq\sqrt{3(a^2+b^2+c^2)}$ is homogeneous, we can assume $b=c=1$, which gives
$$\frac{2(a^2+1)}{a+1}+1\geq\sqrt{3(a^2+2)}$$ or
$$(a-1)^2(a^2+3)\geq0.$$
Done!
