Prove: $\sum\frac{\sqrt{a+3}}{a+\sqrt{bc}}\ge\frac{2(\sqrt{a}+\sqrt{b}+\sqrt{c})}{\sqrt{a+b+c+1}}$

Question

Problem: Let $a,b,c\ge0: ab+bc+ca=3$. Prove that: $$\dfrac{\sqrt{a+3}}{a+\sqrt{bc}}+\dfrac{\sqrt{b+3}}{b+\sqrt{ca}}+\frac{\sqrt{c+3}}{c+\sqrt{ab}}\geq\dfrac{2(\sqrt{a}+\sqrt{b}+\sqrt{c})}{\sqrt{a+b+c+1}}$$

I'm stuck in this tough problem. Here is two approach I tried:

(1): Notice that:$$\dfrac{2(\sqrt{a}+\sqrt{b}+\sqrt{c})}{\sqrt{a+b+c+1}}\leq 3$$ It leads to prove: $$\sqrt{(a+3)(b+3)(c+3)}\geq (a+\sqrt{bc})(b+\sqrt{ac})(c+\sqrt{ab})$$

But the last inequality is not true by checking: $a=b=0,99$

(2): Also by AM-GM, we need to prove: $$27\dfrac{\sqrt{(a+3)(b+3)(c+3)}}{(a+\sqrt{bc})(b+\sqrt{ac})(c+\sqrt{ab})}\ge \left(\dfrac{2(\sqrt{a}+\sqrt{b}+\sqrt{c})}{\sqrt{a+b+c+1}}\right)^3$$ Which seems true but I can not do it due to very high degree. Sorry: this one is also wrong for $a=b=0.01$

Is there any better idea? Or someone can help me full my 2th approach. Thank you

Update: I recieved a nice proof. Hope to see more ideas.

How can we apply the LM or KKT?

Answer 1

Proof.

The proof was inspired by Michael Rozenberg's idea.

Firstly, we use Cauchy-Schwarz as \begin{align*} (a+b+c+1)\sqrt{a+3}&=\sqrt{[a(1+b+c)+bc][4a(1+b+c)+(b+c+1-a)^2]}\\&\ge 2a(1+b+c)+\sqrt{bc}(b+c+1-a), \end{align*} which implies $$(a+b+c+1)\left(\sqrt{a+3}+\sqrt{bc}\right)\ge 2(b+c+1)(a+\sqrt{bc}),$$or$$\frac{\sqrt{a+3}}{a+\sqrt{bc}}\ge \frac{b+c+1-a}{a+b+c+1}+\frac{a}{a+\sqrt{bc}}.\tag{*}$$ Now, sum cyclically on $(*)$ we obtain $$\sum_{cyc}\frac{\sqrt{a+3}}{a+\sqrt{bc}}\ge\frac{a+b+c+3}{a+b+c+1}+\sum_{cyc}\frac{a}{a+\sqrt{bc}}. $$ The last inequality is true by Cauchy-Schwarz and AM-GM and the desired result follows.

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Indeed, by using Cauchy-Schwarz inequality$$\sum_{cyc}\frac{a}{a+\sqrt{bc}}\ge \frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{a+b+c+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}.$$ Also, it is easily $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le 3.$ Hence, apply AM-GM $$\sum_{cyc}\frac{\sqrt{a+3}}{a+\sqrt{bc}}\ge\frac{a+b+c+3}{a+b+c+1}+\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{a+b+c+3}\ge\frac{2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}{\sqrt{a+b+c+1}} . $$ The proof is done. Equality holds at $a=b=c=1.$

Motivation.

The key step is $(*).$

When I tried to multiply $a+b+c+1$ to $LHS$ of the OP, I came up with Cauchy-Schwarz 's idea.