# Given three positve numbers $a,b,c$. Prove that $\sum\limits_{cyc}\frac{a}{\sqrt{b(a+b)}}\geqq \sum\limits_{cyc}\frac{a}{\sqrt{b(c+a)}}$ .

Given three positive numbers $$a, b, c$$, prove that $$\sum\limits_{cyc}\frac{a}{\sqrt{b(a+ b)}}\geqq \sum\limits_{cyc}\frac{a}{\sqrt{b(c+ a)}}.$$ I tried on Holder inequality and https://artofproblemsolving.com/community/c6h194103p1065812 but it's hard! I need to the hints and hope to see the $$uvw$$ help here! Thanks a lot!

• At first glance, this looks to me like something for the rearrangement inequality. Have you tried applying that? May 18, 2019 at 14:28
• @HaiDangel If it's true it's a very interesting problem. Where did you take this inequality? May 19, 2019 at 6:40
• @HaiDangel Where you took this problem? May 20, 2019 at 5:55

The inequality is true. The Buffalo Way works. Here is a hint. Currently, I do not give the Buffalo Way part for people to attempt.

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With the substitutions $$(a, b, c) \to (a^2, b^2, c^2)$$, the inequality becomes $$\sum_{\mathrm{cyc}} \frac{a^2}{b\sqrt{a^2+b^2}} \ge \sum_{\mathrm{cyc}} \frac{a^2}{b\sqrt{c^2+a^2}}.$$

WLOG, assume that $$c = \min(a,b,c) = 1$$.

Let $$X = \frac{a^2+b^2}{2}, Y = \frac{b^2+c^2}{2}, Z = \frac{c^2 + a^2}{2}$$. We need to prove that $$\frac{a^2}{b\sqrt{X}} + \frac{b^2}{c\sqrt{Y}} + \frac{c^2}{a\sqrt{Z}} \ge \frac{a^2}{b\sqrt{Z}} + \frac{b^2}{c\sqrt{X}} + \frac{c^2}{a\sqrt{Y}}$$ or $$\big(\frac{b^2}{c} - \frac{c^2}{a}\big)\frac{1}{\sqrt{Y}} \ge \big(\frac{b^2}{c} - \frac{a^2}{b}\big)\frac{1}{\sqrt{X}} + \big(\frac{a^2}{b} - \frac{c^2}{a}\big)\frac{1}{\sqrt{Z}}$$ Since $$\frac{b^2}{c} - \frac{c^2}{a} \ge 0$$, it suffices to prove that $$\big(\frac{b^2}{c} - \frac{c^2}{a}\big)^2\frac{1}{Y} \ge \Big[\big(\frac{b^2}{c} - \frac{a^2}{b}\big)\frac{1}{\sqrt{X}} + \big(\frac{a^2}{b} - \frac{c^2}{a}\big)\frac{1}{\sqrt{Z}}\Big]^2$$ or $$\big(\frac{b^2}{c} - \frac{c^2}{a}\big)^2\frac{1}{Y} \ge \big(\frac{b^2}{c} - \frac{a^2}{b}\big)^2\frac{1}{X} + \big(\frac{a^2}{b} - \frac{c^2}{a}\big)^2\frac{1}{Z} + 2\big(\frac{b^2}{c} - \frac{a^2}{b}\big)\big(\frac{a^2}{b} - \frac{c^2}{a}\big)\frac{1}{\sqrt{ZX}}.$$ There are three possible cases:

Case I $$b \ge a^3$$: Clearly $$\big(\frac{b^2}{c} - \frac{a^2}{b}\big)\big(\frac{a^2}{b} - \frac{c^2}{a}\big) \le 0$$. Since $$\sqrt{ZX}\le \frac{Z+X}{2}$$, it suffices to prove that $$\big(\frac{b^2}{c} - \frac{c^2}{a}\big)^2\frac{1}{Y} \ge \big(\frac{b^2}{c} - \frac{a^2}{b}\big)^2\frac{1}{X} + \big(\frac{a^2}{b} - \frac{c^2}{a}\big)^2\frac{1}{Z} + 2\big(\frac{b^2}{c} - \frac{a^2}{b}\big)\big(\frac{a^2}{b} - \frac{c^2}{a}\big)\frac{2}{Z+X}.$$ 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 $$14$$. The Buffalo Way works.

Case II $$b \le a^{2/3}$$: Clearly $$\big(\frac{b^2}{c} - \frac{a^2}{b}\big)\big(\frac{a^2}{b} - \frac{c^2}{a}\big) \le 0$$. Since $$\sqrt{ZX}\le \frac{Z+X}{2}$$, it suffices to prove that $$\big(\frac{b^2}{c} - \frac{c^2}{a}\big)^2\frac{1}{Y} \ge \big(\frac{b^2}{c} - \frac{a^2}{b}\big)^2\frac{1}{X} + \big(\frac{a^2}{b} - \frac{c^2}{a}\big)^2\frac{1}{Z} + 2\big(\frac{b^2}{c} - \frac{a^2}{b}\big)\big(\frac{a^2}{b} - \frac{c^2}{a}\big)\frac{2}{Z+X}.$$ After clearing the denominators, it suffices to prove that $$g(a,b,c)\ge 0$$ where $$g(a,b,c)$$ is a homogeneous polynomial of degree $$14$$. The Buffalo Way works.

Case III $$a^{2/3}< b < a^3$$: Clearly $$\big(\frac{b^2}{c} - \frac{a^2}{b}\big)\big(\frac{a^2}{b} - \frac{c^2}{a}\big) \ge 0$$. Since $$\sqrt{ZX} \ge \frac{ac + ab}{2}$$, it suffices to prove that $$\big(\frac{b^2}{c} - \frac{c^2}{a}\big)^2\frac{1}{Y} \ge \big(\frac{b^2}{c} - \frac{a^2}{b}\big)^2\frac{1}{X} + \big(\frac{a^2}{b} - \frac{c^2}{a}\big)^2\frac{1}{Z} + 2\big(\frac{b^2}{c} - \frac{a^2}{b}\big)\big(\frac{a^2}{b} - \frac{c^2}{a}\big)\frac{2}{ab+ac}.$$ After clearing the denominators, it suffices to prove that $$h(a,b,c)\ge 0$$ where $$h(a,b,c)$$ is a homogeneous polynomial of degree $$13$$. The Buffalo Way works.