# 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.

• Looks like a tough nut. ... Feb 9, 2017 at 8:22
• @Andreas Sometimes I think that this inequality is wrong, but I don't see a counterexample. Feb 9, 2017 at 12:35
• Equality occurs also for $(a,b,c) = (0,b,b)$. Further, I can prove that the inequality holds for $(a,b,c) = (0,b,c)$. The remaining case, due to homogeneity, is then e.g. $(a,b,c) = (1,b,c)$. But you don't acknowledge partial proofs .... Feb 9, 2017 at 13:58

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

• Nice. Is it easy to prove the estimation? May 20, 2020 at 12:42
• @River Li I think, it's not so easy, but it solves the problem. May 20, 2020 at 13:40
• (+1) It is actually a nice form. May 20, 2020 at 14:08
• @bamboo_jjvy271 For your values of $a$, $b$ and $c$ the right hand side is negative. May 21, 2020 at 5:45
• Oh my mistake..
– user552223
May 21, 2020 at 5:48

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.

• How did you find $f(x)$? Feb 28, 2021 at 2:51
• @tthnew According to the equality condition $(a, b, c) = (1, 1, 0)$ (correspondingly $X = 36/25$), we use the method of unknown coefficients to find $\sqrt{x} \ge \frac{ax^2 + bx + c}{x^2 + px + q}$ with equality condition $x = 36/25$ etc. Feb 28, 2021 at 2:59
• It was nice solution, +1) Feb 28, 2021 at 3:02
• @tthnew Thanks. The isolated fudging by Nguyenhuyen_AG in another answer is better. Feb 28, 2021 at 3:03
• My software output the same result as Nguyenhuyen_AG for this example! Mar 10, 2021 at 13:10

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.

• Nice again +1. Isolated fudging is powerful Sep 20 at 12:15
• @TATAbox Thanks. Sep 20 at 12:32
• could you please take a look help me? math.stackexchange.com/questions/4774103/… Sep 24 at 10:55
• @TATAbox Currently I have no idea. I upvoted yesterday. I don't know why two downvotes. Sep 24 at 11:41
• It doesn't matter but thank you anyway. Does isolated fuding technique work? I think it should be unique way to kill two equality cases. Sep 24 at 11:58