# Prove $\sqrt{\frac{a+4bc}{4a+bc}}+\sqrt{\frac{b+4ca}{4b+ac}}+\sqrt{\frac{c+4ab}{4c+ab}}\ge 3,$ when $a+b+c=3.$

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.

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 $$0, and we need to prove that: $$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!

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