<s>[INST] 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?
 [/INST] 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.
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