Prove that $\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}+\frac{a^2+b^2}{a+b} \ge a+b+c$ If $a,b,c$ are positive , show that $$\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b} \ge a+b+c$$
Trial: Here  I proceed in this way $$\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b} \ge  \dfrac{2bc}{b+c}+\dfrac{2ca}{c+a}+\dfrac{2ab}{a+b}$$ then how I proceed. Please help.
 A: Potato has already answered this question elegantly. Here is another solution.
By Cauchy-Schwarz, we get
$$((b+c)+(c+a)+(a+b))\left(\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}+\frac{a^2+b^2}{a+b}\right)\ge \left(\sqrt{b^2+c^2}+\sqrt{c^2+a^2}+\sqrt{a^2+b^2}\right)^2 $$
To prove the original inequality, it suffices to show that
$$\left(\sqrt{b^2+c^2}+\sqrt{c^2+a^2}+\sqrt{a^2+b^2}\right)^2 \ge 2(a+b+c)^2$$
which boils down to showing
$$\sqrt{a^2+b^2}\sqrt{b^2+c^2} + \sqrt{c^2+a^2}\sqrt{a^2+b^2}+\sqrt{b^2+c^2}\sqrt{c^2+a^2} \ge 2(ab+bc+ca)$$
With one more application of Cauchy-Schwarz, 
$$\begin{align}
(b^2+a^2)(b^2+c^2)\ge (b^2+ac)^2 \\
(c^2+b^2)(c^2+a^2)\ge (c^2+ba)^2 \\
(a^2+b^2)(a^2+c^2)\ge (a^2+bc)^2
\end{align}$$
Taking square roots of both sides, and adding them up, it suffices to prove
$$ a^2+b^2+c^2\ge ab+bc+ca$$
which is equivalent to 
$$ (a-b)^2+(b-c)^2+(c-a)^2\ge 0$$
So we are done :) I agree that it is a bit overkill. But I like Cauchy-Schwarz inequality. 
A: We have by Titu's Lemma, for any $x,y$ reals and $a,b >0$
$ \dfrac{x^2}{a}+\dfrac{y^2}{b} \ge \dfrac{(x+y)^2}{a+b}$
$\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b} \ge a+b+c$ is quite direct. :)
A: It suffices to show, by symmetry, that 
$$\frac{a^2+b^2}{a+b}\ge \frac{a+b}{2}.$$
This is equivalent to 
$$2a^2+2b^2\ge a^2+b^2+2ab\iff a^2+b^2\ge 2ab.$$
The last inequality is just the standard AM-GM inequality. 
