A geometric inequality for acute triangle $\sum \sqrt \frac{b^2+c^2-a^2}{a^2+2bc} \leq \sqrt3$ I am trying to prove $\sum \sqrt \frac{b^2+c^2-a^2}{a^2+2bc} \leq \sqrt3$ for an acute triangle. There is:
$$\sum \sqrt \frac{b^2+c^2-a^2}{a^2+2bc} \leq \sqrt{3\sum\frac{b^2+c^2-a^2}{a^2+2bc}}$$
So we have to prove: $\sum\frac{b^2+c^2-a^2}{a^2+2bc}\leq 1 $
We observe that $\frac{b^2+c^2-a^2}{a^2+2bc}=\frac{(b+c)^2}{a^2+2bc}-1$ and I can't go further

We can end the beautiful Atticus's idea also by the following C-S:
$$\sum_{cyc}\frac{x}{\sqrt{\frac{y^2+z^2}{2}+x^2+yz}}=\sum_{cyc}\frac{x\sqrt6}{\sqrt{(1+2)(2x^2+(y+z)^2}}\leq\sum_{cyc}\frac{x\sqrt6}{\sqrt2x+\sqrt2(y+z)}=\sqrt3.$$
