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Problem. If $a,b,c\ge 0: ab+bc+ca>0$ and $a+b+c=3,$ prove that$$\sqrt{\frac{a+4bc}{4a+bc}}+\sqrt{\frac{b+4ca}{4b+ac}}+\sqrt{\frac{c+4ab}{4c+ab}}\ge 3.$$

It was here.

Equality holds at $a=b=c=1$ and $abc=0.$

I've tried to use AM-GM but the following inequality is not true$$\sqrt{\frac{a+4bc}{4a+bc}}\sqrt{\frac{b+4ca}{4b+ac}}\sqrt{\frac{c+4ab}{4c+ab}}\le 1.$$

I hope Isolated fudging method helps. Indeed, we'll prove $$\sum_{cyc}\sqrt{\frac{\dfrac{a(a+b+c)}{3}+4bc}{\dfrac{4a(a+b+c)}{3}+bc}}\ge 3.$$Or$$\sum_{cyc}\sqrt{\frac{a^2+ab+12bc+ca}{4a^2+4ab+3bc+4ca}}\ge 3.$$ I failed to use it. Hope you give me a hint to kill this problem. Thank you.

Michael Rozenberg
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Tran Ngoc Khuong Trang
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1 Answers1

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If $abc=0$, then it's an equality.

Let $abc \neq 0$, $\frac{bc}{a} = 3\tan^2\frac{\alpha}{2}$, $\frac{ac}{b} = 3\tan^2\frac{\beta}{2}$, and $\frac{ab}{c} = 3\tan^2\frac{\gamma}{2}$, where $\{\alpha, \beta, \gamma\} \subset (0, \pi)$.

Thus, the condition gives $$ \tan\frac{\alpha}{2}\tan\frac{\beta}{2} + \tan\frac{\alpha}{2}\tan\frac{\gamma}{2} + \tan\frac{\beta}{2}\tan\frac{\gamma}{2} = 1, $$ which implies $$ \alpha + \beta + \gamma = \pi, $$ and we need to prove that $\sum\limits_{\text{cyc}} f(\alpha) \geq 3$, where $$ f(x) = \sqrt{\frac{12\tan^2\frac{x}{2} + 1}{3\tan^2\frac{x}{2} + 4}}. $$ But, $$ f''(x) = \frac{45(44\cos^3 x - 90\cos^2 x + 265\cos x - 76)}{4\sqrt{(13 - 11\cos x)^3 (7 + \cos x)^5}}, $$ which shows that $f$ has a unique inflection point on $(0, \pi)$. By Vasc's HCF Theorem, it is enough to prove $\sum\limits_{\text{cyc}} f(\alpha) \geq 3$ for the equality case of two variables.

For more on HCF, see: Vasile Cirtoaje, "Mathematical Inequalities", 2018, Volume 4, page 3.

Now, we can complete the proof as follows.

Let $\frac{ab}{c} = z$, $\frac{ac}{b} = y$, and $\frac{bc}{a} = x$.

Thus, the condition gives: $$ \sqrt{xy} + \sqrt{xz} + \sqrt{yz} = 3, $$ and we need to prove that: $$ \sum_{\text{cyc}} \sqrt{\frac{4x + 1}{x + 4}} \geq 3 $$ for the equality case of two variables.

Let $y = x$. Thus, $z = \frac{(3 - x)^2}{4x}$, where $0 < x < 3$, and we need to prove that: $$ 2\sqrt{\frac{4x + 1}{x + 4}} + \sqrt{\frac{4\left(\frac{3 - x}{2\sqrt{x}}\right)^2 + 1}{\left(\frac{3 - x}{2\sqrt{x}}\right)^2 + 4}} \geq 3, $$ or equivalently, $$ 2\sqrt{\frac{4x + 1}{x + 4}} + 2\sqrt{\frac{x^2 - 5x + 9}{x^2 + 10x + 9}} \geq 3. $$ After squaring both sides, we obtain: $$ 11x^3 + 34x^2 - 301x - 144 + 8\sqrt{(4x + 1)(x + 4)(x^2 - 5x + 9)(x^2 + 10x + 9)} \geq 0, $$ which is true by the Cauchy-Schwarz inequality and AM-GM: $$ 11x^3 + 34x^2 - 301x - 144 + 8\sqrt{(4x + 1)(x + 4)(x^2 - 5x + 9)(x^2 + 10x + 9)} = $$ $$ = 11x^3 + 34x^2 - 301x - 144 + 8\sqrt{((2x + 2)^2 + 9x)((3 - x)^2 + x)((3 + x)^2 + 4x)} \geq $$ $$ \geq 11x^3 + 34x^2 - 301x - 144 + 8\left(2x + 2 + \frac{2x}{x + 1}\right)(9 - x^2 + 2x) = $$ $$ = \frac{x(x - 1)^2(19 - 5x)}{x + 1} \geq 0, $$ and we are done!

Wade
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Michael Rozenberg
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