Calculate the minimum value of $\sum_\mathrm{cyc}\frac{a^2}{b + c}$ where $a, b, c > 0$ and $\sum_\mathrm{cyc}\sqrt{a^2 + b^2} = 1$. 
$a$, $b$ and $c$ are positives such that $\sqrt{a^2 + b^2} + \sqrt{b^2 + c^2} + \sqrt{c^2 + a^2} = 1$. Calculate the minimum value of $$\frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b}$$

I have not come up with any ideas to solve the problem yet. I will probably in  the near future but right now, I can't.

By rearrangement-inequality
$$\frac{a^{\,2}}{b+ c}+ \frac{b^{\,2}}{c+ a}+ \frac{c^{\,2}}{a+ b}\geqq \frac{a^{\,2}}{a+ b}+ \frac{b^{\,2}}{b+ c}+ \frac{c^{\,2}}{c+ a}$$
We have to prove
$$\frac{a^{\,2}}{a+ b}+ \frac{b^{\,2}}{b+ c}+ \frac{c^{\,2}}{c+ a}\geqq \frac{1}{2}(\,\sqrt{\frac{a^{\,2}+ b^{\,2}}{2}}+ \sqrt{\frac{b^{\,2}+ c^{\,2}}{2}}+ \sqrt{\frac{c^{\,2}+ a^{\,2}}{2}}\,)$$
By a.m.-g.m.-inequality
$$\sqrt{\frac{a^{\,2}+ b^{\,2}}{2}}= \frac{2\sqrt{2(\,a^{\,2}+ b^{\,2}\,)(\,a+ b\,)^{\,2}}}{4(\,a+ b\,)}\leqq \frac{(\,a+ b\,)^{\,2}+ 2(\,a^{\,2}+ b^{\,2}\,)}{4(\,a+ b\,)}$$
We need to have
$$\frac{4\,a^{\,2}}{a+ b}+ \frac{4\,b^{\,2}}{b+ c}+ \frac{4\,c^{\,2}}{c+ a}\geqq \frac{1}{2}\times \sum\limits_{cyc}\frac{(\,a+ b\,)^{\,2}+ 2(\,a^{\,2}+ b^{\,2}\,)}{(\,a+ b\,)}$$
$$\because\,\frac{a^{\,2}- 2\,ab+ b^{\,2}}{a+ b}+ \frac{b^{\,2}- 2\,bc+ c^{\,2}}{b+ c}+ \frac{c^{\,2}- 2\,ca+ a^{\,2}}{c+ a}\geqq 0$$
So it holds !
$$\therefore\,\frac{a^{\,2}}{b+ c}+ \frac{b^{\,2}}{c+ a}+ \frac{c^{\,2}}{a+ b}\geqq \frac{1}{2}(\,\sqrt{\frac{a^{\,2}+ b^{\,2}}{2}}+ \sqrt{\frac{b^{\,2}+ c^{\,2}}{2}}+ \sqrt{\frac{c^{\,2}+ a^{\,2}}{2}}\,)= \frac{1}{2\sqrt{2}}$$
