Prove that $\sum\limits_{cyc}\sqrt{\frac{8ab+8ac+9bc}{(2b+c)(b+2c)}}\geq5$

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

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.