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!
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1$\begingroup$ At first glance, this looks to me like something for the rearrangement inequality. Have you tried applying that? $\endgroup$– ArthurMay 18, 2019 at 14:28
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1$\begingroup$ @HaiDangel If it's true it's a very interesting problem. Where did you take this inequality? $\endgroup$– Michael RozenbergMay 19, 2019 at 6:40
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1$\begingroup$ @HaiDangel Where you took this problem? $\endgroup$– Michael RozenbergMay 20, 2019 at 5:55
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1 Answer
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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.