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.

I have an alternative solution, by CS:
$$A=\sum\limits_{cyc} \frac{a^2}{b+c}=\sum\limits_{cyc} \frac{a^4}{a^2b+a^2c}\geq \frac{\left(\sum\limits_{cyc}a^2\right)^2}{\sum\limits_{sym}a^2b} $$ 
By Jensen: 
$$1=\sum\limits_{cyc}\sqrt{a^2+b^2}\leq 3\sqrt{\frac{2}{3}(a^2+b^2+c^2)}$$
So 
$$A\geq \frac{\left(\sum\limits_{cyc}a^2\right)^2}{(\sum\limits_{sym}a^2b )* 1}\geq \frac{1}{\sqrt{6}}\frac{\left(\sum\limits_{cyc}a^2\right)^{\frac{3}{2}}}{\sum\limits_{sym}a^2b}$$
If you plug in $a=b=c=\frac{1}{3\sqrt{2}}$ you will get $A=\frac{1}{2\sqrt{2}}$
We will show that $\frac{1}{\sqrt{6}}\frac{\left(\sum\limits_{cyc}a^2\right)^{\frac{3}{2}}}{\sum\limits_{sym}a^2b}\geq \frac{1}{2\sqrt{2}}$ 
Or
$$4\left(\sum\limits_{cyc}a^2\right)^{3}\geq 3\left(\sum\limits_{sym}a^2b\right)^2$$
Which is true after foiling by Muirhead and Schur
