Prove that $\sum\limits_{cyc}\sqrt{\frac{8ab+8ac+9bc}{(2b+c)(b+2c)}}\geq5$ Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that:
$$\sqrt{\frac{8ab+8ac+9bc}{(2b+c)(b+2c)}}+\sqrt{\frac{8ab+8bc+9ac}{(2a+c)(a+2c)}}+\sqrt{\frac{8ac+8bc+9ab}{(2a+b)(a+2b)}}\geq5$$
I tried Holder, but without success.
For example, Holder even with $(ka^2+b^2+c^2+nab+nac+mbc)^3$ does not help.
SOS or C-S seems very ugly here.
The equality occurs also for $(a,b,c)=(1,1,0)$, which adds a problems.

There is the following estimation of Nguyenhuyen_AG.
$$\sqrt{\frac{8a(b+c)+9bc}{(2b+c)(2c+b)}} \geqslant\frac{17a^2-b^2-c^2+20(ab+ac+bc)}{3(a^2+b^2+c^2+4(ab+bc+ca))}$$
