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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.

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?

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

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.