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?
 A: 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
$$6 + \frac12\sum_{cyc}(s-c)c \ge 3s \iff 6 + \frac{s^2}2-\frac32 \ge 3s \iff (s-3)^2\ge 0$$
A: 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!
A: Applying AM-GM, we get:
$\dfrac{\dfrac{a^2 + b^2}{a+b} + \dfrac{a^2 + c^2}{a+c} + \dfrac{c^2 + b^2}{c+b}}{3} \geq \Bigg( \Bigg(\dfrac{a^2 + b^2}{a+b} \Bigg) \Bigg(\dfrac{a^2 + c^2}{a+c} \Bigg) \Bigg(\dfrac{c^2 + b^2}{c+b} \Bigg)\Bigg)^{\frac{1}{3}}$
$\geq \Bigg( \Bigg(\dfrac{a+ b}{2} \Bigg) \Bigg(\dfrac{a+ c}{2} \Bigg) \Bigg(\dfrac{c + b}{2} \Bigg)\Bigg)^{\frac{1}{3}}$
(Follows from AM-GM again.)
$\geq \frac{1}{2} ((a+b)(b+c)(a+c))^{\frac{1}{3}}$
$\geq \frac{1}{2} ((a+b+c)(ab+bc+ca) -abc)^{\frac{1}{3}} \quad \ldots (1)$ 
Clearly $(abc)_{max} = \bigg( \dfrac{a^2 + b^2 +c^2}{3}\bigg)^{\frac{3}{2}} = 1$
Now by Lagrange's method of undetermined multipliers it isn't hard to show that subject to the constraint $a^2 +b^2 +c^2 = 3$, the maximum values of $a+b+c$ and $ab +bc +ca$ are both 3.
Together we get:
$\dfrac{\dfrac{a^2 + b^2}{a+b} + \dfrac{a^2 + c^2}{a+c} + \dfrac{c^2 + b^2}{c+b}}{3} \geq \dfrac{1}{2}((3 \times 3) -1)^{\frac{1}{3}}$
And the required inequality follows.
