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.
 A: There is the following estimation of Nguyenhuyen_AG.
$$\sqrt{\frac{8a(b+c)+9bc}{(2b+c)(2c+b)}} \geqslant\frac{17a^2-b^2-c^2+20(ab+ac+bc)}{3(a^2+b^2+c^2+4(ab+bc+ca))}$$
A: The Buffalo Way works. Due to symmetry, assume that $a\ge b\ge c$.
Let 
$$X = \frac{9}{25}\frac{8ab + 9bc + 8ca}{(2b+c)(b+2c)}, \quad
Y = \frac{9}{25}\frac{8bc + 9ca + 8ab}{(2c+a)(c+2a)}, \quad
Z = \frac{9}{25}\frac{8ca + 9ab + 8bc}{(2a+b)(a+2b)}.$$
We need to prove that $\sqrt{X} + \sqrt{Y} + \sqrt{Z} \ge 3$.
We will use the following bounds:
$$\sqrt{x} \ge f(x) = \frac{22x(5x+6)}{25x^2 + 181x+36}, \quad \forall x \ge 0$$
and
$$\sqrt{x} \ge g(x) = \frac{16x(5x+3)}{25x^2+94x+9}, \quad \forall x\ge 0$$
since
$$x - \Big(\frac{22x(5x+6)}{25x^2 + 181x+36}\Big)^2 = \frac{x(x-1)^2(25x-36)^2}{(25x^2 + 181x+36)^2}$$
and
$$x - \Big(\frac{16x(5x+3)}{25x^2+94x+9}\Big)^2 =  \frac{x(25x-9)^2(x-1)^2}{(25x^2 + 94x + 9)^2}.$$
It suffices to prove that $f(X) + f(Y) + g(Z) \ge 3$.
After clearing the denominators, it suffices to prove that $F(a,b,c)\ge 0$
where $F(a,b,c)$ is a homogeneous polynomial of degree $12$. 
If $c = 0$, we have
\begin{align}
F(a,b,0) &= 64a^2b^2(a-b)^2\left(52500a^6+1729775a^5b+13521612a^4b^2\right.\\
&\quad \left. +27797474a^3b^3+13521612a^2b^4+1729775ab^5+52500b^6\right).
\end{align}
The inequality is true.
If $c > 0$, let $c = 1, \ b = 1+s, \ a = 1 + s + t$ for $s, t \ge 0$,
then $F(1+s+t, 1+s, 1)$ is a polynomial in $s, t$ with non-negative coefficients. The inequality is true. 
We are done. 
