<s>[INST] $\frac{2-a}{a+\sqrt{bc+abc}}+\frac{2-b}{b+\sqrt{ca+abc}}+\frac{2-c}{c+\sqrt{ab+abc}}\ge1$ 
For $a,b,c\ge0: ab+bc+ca+abc=4$ then: $$\frac{2-a}{a+\sqrt{bc+abc}}+\frac{2-b}{b+\sqrt{ca+abc}}+\frac{2-c}{c+\sqrt{ab+abc}}\ge1$$

I used the condition and get: $a+\sqrt{bc+abc}=a+\sqrt{4-a(b+c)}\le a+2$
So we need to prove that: $$\frac{2-a}{a+2}+\frac{2-b}{b+2}+\frac{2-c}{c+2}\ge1$$
I tried to full expand but the rest seems complicated for me.
Can anyone help me full my idea? Every thinking is welcomed, thanks!
 [/INST] Now I find a proof by AM-GM: $$4=2\sqrt{\frac{a(b+c)+bc+abc}{a+\sqrt{bc+abc}}(a+\sqrt{bc+abc})}\le a+\sqrt{bc+abc}+\frac{a(b+c)+bc+abc}{a+\sqrt{bc+abc}}=\frac{a(a+b+c)}{a+\sqrt{bc+abc}}+2\sqrt{bc+abc}$$
$$\implies \frac{2+\sqrt{bc+abc}}{a+\sqrt{bc+abc}}\ge\frac{2(b+c)}{a+b+c}$$
Or: $$\frac{2-a}{a+\sqrt{bc+abc}}\ge\frac{b+c-a}{a+b+c}$$
Sum up similar inequalities, we get desired result!
Equality holds iff $(a,b,c)=(0,2,2)$ and pers
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