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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?

Anonymous
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    https://artofproblemsolving.com/community/c6t243f6h2755079_sqrt_inequality – Anonymous Oct 16 '22 at 02:11
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    I feel the only hope to find an answer to this post is applying Lagrange multipliers. – Piquito Oct 16 '22 at 17:40
  • Please excuse more help me. How to apply Lagrange multipliers? – Anonymous Oct 17 '22 at 04:30
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    It would be difficult with your function. The method is "standard" and easy but the algebraic system at the end (in your case with four unknowns) will be not easy. Search in Google Lagrange multipliers and you will find many good references. (Sorry for bad English). – Piquito Oct 17 '22 at 13:04
  • I try a lot. It is impossible hand- solution! – Anonymous Oct 20 '22 at 07:50
  • Thuong: Note that when un of $a,b,c$ is large the other two should be small and the inequality can be deduced without difficulty. The problem is with the small values at neighborhood $1$ because all the three values cannot be smaller than $1$. – Piquito Oct 21 '22 at 13:12
  • I found a good way by AM-GM, magically – Anonymous Oct 23 '22 at 09:53
  • Thuong: Post your answer yourself, it would be well received. – Piquito Oct 24 '22 at 11:22

1 Answers1

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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.

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.

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