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.

SOS helps!
Let $a=b=c=\frac{1}{3\sqrt2}.$
Thus, we get a value $\frac{1}{2\sqrt2}.$
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$\sum_{cyc}\frac{a^2}{b+c}\geq\frac{1}{2\sqrt2}\sum_{cyc}\sqrt{a^2+b^2}$$ or
$$\sum_{cyc}\left(\frac{a^2}{b+c}-\frac{a}{2}\right)\geq\frac{1}{4}\sum_{cyc}\left(\sqrt{2(a^2+b^2)}-a-b\right)$$ or
$$\sum_{cyc}\frac{a(a-b-(c-a))}{b+c}\geq\frac{1}{2}\sum_{cyc}\frac{(a-b)^2}{\sqrt{2(a^2+b^2)}+a+b}$$ or
$$\sum_{cyc}(a-b)\left(\frac{a}{b+c}-\frac{b}{a+c}\right)\geq\frac{1}{2}\sum_{cyc}\frac{(a-b)^2}{\sqrt{2(a^2+b^2)}+a+b}$$ or
$$\sum_{cyc}\frac{(a-b)^2(a+b+c)}{(a+c)(b+c)}\geq\frac{1}{2}\sum_{cyc}\frac{(a-b)^2}{\sqrt{2(a^2+b^2)}+a+b}$$ and since by C-S $$\sqrt{2(a^2+b^2)}=\sqrt{(1^2+1^2)(a^2+b^2)}\geq a+b,$$ it's enough to prove that:
$$\sum_{cyc}\frac{(a-b)^2(a+b+c)}{(a+c)(b+c)}\geq\frac{1}{4}\sum_{cyc}\frac{(a-b)^2}{a+b}$$ or
$$\sum_{cyc}(a-b)^2(4(a+b+c)(a+b)-(a+c)(b+c))\geq0,$$ for which it's enough to prove that
$$\sum_{cyc}(a-b)^2(a^2+b^2-c^2)\geq0.$$
Now, let $a\geq b\geq c$.
Thus, 
 $$\sum_{cyc}(a-b)^2(a^2+b^2-c^2)\geq(a-c)^2(a^2+c^2-b^2)+(b-c)^2(b^2+c^2-a^2)\geq$$
$$\geq(b-c)^2(a^2-b^2)+(b-c)^2(b^2-a^2)=0$$ and we are done!
