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

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

My second proof.

In Michael Rozenberg's answer, Nguyenhuyen_AG's idea is the so-called isolated fudging.

I got the same form using the isolated fudging technique.

We need to prove that $$\sqrt{\frac{8ab + 8ca + 9bc}{(2b + c)(2c + b)}} \ge \frac{17a^2 + 20(ab + bc + ca) - b^2 - c^2}{3(a^2 + b^2 + c^2) + 12(ab + bc + ca)}. \tag{1}$$

We split into two cases.

Case 1: $a = 0$

WLOG, assume that $b = 1$. It suffices to prove that $$\sqrt{\frac{9c}{(2 + c)(2c + 1)}} \ge \frac{20c - 1 - c^2}{3(1 + c^2) + 12c}. \tag{2}$$ (2) is true.

Case 2: $a > 0$

WLOG, assume that $a = 1$. Let $p = b + c, q = bc$. We have $p^2 \ge 4q$.

(1) is written as $$\sqrt{\frac{8p + 9q}{2p^2 + q}} \ge \frac{-p^2 + 20p + 22q + 17}{3p^2 + 12p + 6q + 3}. \tag{3}$$

We only need to prove the case that $$-p^2 + 20p + 22q + 17 \ge 0$$ or $$q \ge \frac{p^2 - 20p - 17}{22}. \tag{4}$$

From (3) and (4), it suffices to prove that $$\frac{8p + 9q}{2p^2 + q} \ge \left(\frac{-p^2 + 20p + 22q + 17}{3p^2 + 12p + 6q + 3}\right)^2$$ or \begin{align*} f(q)&:= -160\,{q}^{3}+ \left( -600\,{p}^{2}+704\,p-424 \right) {q}^{2}\\ &\qquad + \left( 168\,{p}^{4}-784\,{p}^{3}+748\,{p}^{2}+256\,p-208 \right) q\\ &\qquad -2 \,{p}^{6}+152\,{p}^{5}-156\,{p}^{4}-64\,{p}^{3}-2\,{p}^{2}+72\,p\\ &\ge 0. \end{align*}

We have $$f''(q) = -1200p^2 + 1408p - 960q - 848 \le -1200p^2 + 1408p - 848 < 0.$$ Thus, $f(q)$ is concave. Also, we have $f(\frac{p^2 - 20p - 17}{22}) \ge 0$ and $f(p^2/4) \ge 0$ (easy). Thus, we have $f(q) \ge 0$ for all $q\in [\frac{p^2 - 20p - 17}{22}, p^2/4]$.

We are done.