<s>[INST] 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.
 [/INST] By rearrangement inequality, 
$$\sum_{cyc} \frac{a^2}{b+c} \ge \sum_{cyc} \frac{b^2}{b+c}.\tag{1}$$
So
$$2\sum_{cyc}\frac{a^2}{b+c} \ge \sum_{cyc}\frac{a^2+b^2}{b+c}.\tag{2}$$
Now, by C-S
$$\sum_{cyc}(b+c) \sum_{cyc}\frac{a^2+b^2}{b+c} \ge \left(\sum\sqrt{a^2+b^2}\right)^2.\tag{3}$$
It's also clear that
$$\sum_{cyc}(b+c) \le \sqrt2 \sum_{cyc}\sqrt{a^2+b^2}\tag{4}.$$
From (2), (3), and (4), we conclude that $\sum \frac{a^2}{b+c}$ attains its minimum at $a=b=c$.
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