$\frac{2-a}{a+\sqrt{bc+abc}}+\frac{2-b}{b+\sqrt{ca+abc}}+\frac{2-c}{c+\sqrt{ab+abc}}\ge1$ 
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!
 A: Let $x = a + 2, \; y = b + 2, \; z = c + 2$
Hence, we need to prove that
$$\frac{4-x}{x} + \frac{4-y}{y} + \frac{4-z}{z} \geq 1$$
$$\implies \frac{4}{x} - \frac{x}{x} + \frac{4}{y} - \frac{y}{y} + \frac{4}{z} - \frac{z}{z} \geq 1$$
$$\implies \frac{4}{x} + \frac{4}{y} + \frac{4}{z} - 3 \geq 1$$
$$\implies \frac{4}{x} + \frac{4}{y} + \frac{4}{z} \geq 4$$
$$\implies \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq 1$$
$$\implies \frac{1}{a+2} + \frac{1}{b+2} + \frac{1}{c+2} \geq 1$$
$$\implies \frac{4(a + b + c) + ab + bc + ca + 12}{4(a + b + c) + 2(ab + bc + ca) + abc + 8} \geq 1$$
$$\implies 4(a + b + c) + ab + bc + ca + 12 \geq 4(a + b + c) + 2(ab + bc + ca) + abc + 8$$
$$\implies ab + bc + ca +abc -4 \leq 0$$
We know that $ab + bc + ca + abc = 4 \implies ab + bc + ca +abc - 4 
 = 4 - 4 = 0$.
A: 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
A: Futhermore, I tried C-S inequality and saw something "tricky":
It is obvious to get equivalent inequality: $$\frac{2+\sqrt{bc+abc}}{a+\sqrt{bc+abc}}+\frac{2+\sqrt{bc+abc}}{a+\sqrt{bc+abc}}+\frac{2+\sqrt{ca+abc}}{b+\sqrt{ac+abc}}\ge4$$
By C-S: $$2(a+b+c)=\sqrt{\left(a(b+c)+bc+abc\right)\left((b+c-a)^2+4a(b+c)\right)}\ge 2a(b+c)+\sqrt{bc+abc}(b+c-a)$$
Or: $$\frac{2+\sqrt{bc+abc}}{a+\sqrt{bc+abc}}\ge \frac{(a+b+c)\sqrt{bc+abc}+2a(b+c)+\sqrt{bc+abc}(b+c-a)}{(a+b+c)(a+\sqrt{bc+abc})}=\frac{2(b+c)}{a+b+c}$$
The rest is similar to above solution.
