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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.

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1 Answer 1

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

Let $abc\neq0$, $\frac{bc}{a}=3\tan^2\frac{\alpha}{2},$ $\frac{ac}{b}=3\tan^2\frac{\beta}{2}$ 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 says $$\alpha+\beta+\gamma=\pi$$ and we need to prove that $\sum\limits_{cyc}f(\alpha)\geq3,$ where $$f(x)=\sqrt{\frac{12\tan^2\frac{x}{2}+1}{3\tan^2\frac{x}{2}+4}}.$$ But, $$f''(x)=\frac{45(44\cos^3x-90\cos^2x+265\cos{x}-76)}{4\sqrt{(13-11\cos{x})^3(7+\cos{x})^5}},$$ which gives that $f$ has an unique inflection point on $(0,\pi)$ and by the Vasc's HCF Theorem it's enough to prove $\sum\limits_{cyc}f(\alpha)\geq3$ for equality case of two variables.

About HCF see: Vasile Cirtoaje "Mathematical inequalities",2018, Volume 4, page 3.

Now, we can end the proof by the following way.

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: $$\sum_{cyc}\sqrt{\frac{4x+1}{x+4}}\geq3$$ for 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: $$2\sqrt{\frac{4x+1}{x+4}}+\sqrt{\frac{4\left(\frac{3-x}{2\sqrt{x}}\right)^2+1}{\left(\frac{3-x}{2\sqrt{x}}\right)^2+4}}\geq3$$ or $$2\sqrt{\frac{4x+1}{x+4}}+2\sqrt{\frac{x^2-5x+9}{x^2+10x+9}}\geq3$$ or after squaring of the both sides $$11x^3+34x^2-301x-144+8\sqrt{(4x+1)(x+4)(x^2-5x+9)(x^2+10x+9)}\geq0,$$ which is true by C-S and AM-GM: $$11x^3+34x^2-301x-144+8\sqrt{(4x+1)(x+4)(x^2-5x+9)(x^2+10x+9)}=$$ $$=11x^3+34x^2-301x-144+8\sqrt{((2x+2)^2+9x)((3-x)^2+x)((3+x)^2+4x)}\geq$$

$$\geq11x^3+34x^2-301x-144+8\left(2x+2+\frac{2x}{x+1}\right)(9-x^2+2x)=$$ $$=\frac{x(x-1)^2(19-5x)}{x+1}\geq0$$ and we are done!

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    $\begingroup$ The substitution is nice. $\endgroup$
    – River Li
    Oct 28, 2023 at 12:08
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    $\begingroup$ @MichaelRozenberg Could you please full it? Thanks. $\endgroup$
    – Anonymous
    Oct 31, 2023 at 1:13
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    $\begingroup$ @Anonymous I added something. See now. $\endgroup$ Oct 31, 2023 at 5:00

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