$\sqrt{a+b+\frac{c}{ab}}+\sqrt{b+c+\frac{a}{bc}}+\sqrt{c+a+\frac{b}{ca}}+\sqrt{abc}\left(\frac{3}{a+b+c}-2\right)\ge\frac{a+b+c}{\sqrt{abc}}$

Question

Problem: Prove that the following inequality: $$\sqrt{a+b+\frac{c}{ab}}+\sqrt{b+c+\frac{a}{bc}}+\sqrt{c+a+\frac{b}{ca}}+\sqrt{abc}\left(\frac{3}{a+b+c}-2\right)\ge\frac{a+b+c}{\sqrt{abc}}$$ holds for all positive real numbers such that: $ab+bc+ca=2(a+b+c)$

It is quite hard to me. I guess equality holds iff: a=b=c=2

I tried to use C-S inequality as : $\sum{\sqrt{a+b+\frac{c}{ab}}}\ge\sqrt{2(\sqrt{a}+\sqrt{b}+\sqrt{c})^2+\frac{(a+b+c)^2}{abc}}$

The rest is quite complicated. Please help me to solve problem. Thanks!

Answer 1

Proof.

Multiply the OP by $ab+bc+ca$ and using given hypothesis, we will prove $$(ab+bc+ca)\sum_{cyc}\sqrt{a+b+\frac{c}{ab}}+2(a+b+c)\sqrt{abc}\left(\frac{3}{a+b+c}-2\right)\ge\frac{(a+b+c)(ab+bc+ca)}{\sqrt{abc}},$$ or $$(ab+bc+ca)\sum_{cyc}\sqrt{a+b+\frac{c}{ab}}\ge\frac{a^2(b+c)+b^2(c+a)+c^2(a+b)-3abc}{\sqrt{abc}}+4(a+b+c)\sqrt{abc}.$$ From now, we can apply Michael Rozenberg's CBS using.

Can you end it now?

Indeed,\begin{align*} (ab+bc+ca)\sqrt{a+b+\frac{c}{ab}}&=\sqrt{\left(a+b+\frac{c}{ab}\right)[4abc(a+b)+(ca+cb-ab)^2]}\\&\ge 2\sqrt{abc}(a+b)+(ca+cb-ab)\sqrt{\frac{c}{ab}}\\&=2\sqrt{abc}(a+b)+\frac{c^2(a+b)-abc}{\sqrt{abc}}.\tag{1} \end{align*} Sum cyclically on $(1)$ and we get desired result.

Equality holds at $a=b=c=2.$