A geometric inequality for acute triangle $\sum \sqrt \frac{b^2+c^2-a^2}{a^2+2bc} \leq \sqrt3$

Question

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

Answer 1

Let $x=\sqrt{b^2+c^2-a^2}, y=\sqrt{c^2+a^2-b^2}$ and $z=\sqrt{a^2+b^2-c^2}$ (they are positive real numbers due to acute-angled restriction). Then:

$$a=\sqrt{\frac{y^2+z^2}{2}},\ b=\sqrt{\frac{z^2+x^2}{2}},\ c=\sqrt{\frac{x^2+y^2}{2}}$$

and the inequality is equivalent with:

$$\sum_{cyc}\frac{x}{\sqrt{\frac{y^2+z^2}{2}+\sqrt{(x^2+y^2)(z^2+x^2)}}}\leq \sqrt{3}$$

Notice that using Cauchy-Schwarz, we have $(x^2+y^2)(x^2+z^2)\geq (x^2+yz)^2$. Thus, it will be enough to prove:

$$\sum_{cyc}\frac{x}{\sqrt{2x^2+(y+z)^2}}\leq \sqrt{\frac{3}{2}}$$

This inequality is homogeneous so normalize with $x+y+z=3$:

$$\sum_{cyc}\frac{x}{\sqrt{2x^2+(3-x)^2}} \leq \sqrt{\frac{3}{2}}$$

However, $2x^2+(3-x)^2=6+3(x-1)^2\geq 6$, and hence:

$$\sum_{cyc}\frac{x}{\sqrt{2x^2+(3-x)^2}} \leq \frac{x+y+z}{\sqrt{6}}=\sqrt{\frac{3}{2}}$$

Answer 2

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