Find minimum of $\sqrt{a}+\sqrt{b}+\sqrt{c}$, given that $a,b,c \ge 0$ and $ab+bc+ca+abc=4$ 
Find the minimum of $\sqrt{a}+\sqrt{b}+\sqrt{c}$, given that $a,b,c \ge 0$ and $ab+bc+ca+abc=4$.

I can prove that $\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=1,$ because:
$$ab+bc+ca+abc=4 \\ \implies (a+2)(b+2)(c+2)=(a+2)(b+2)+(b+2)(c+2)+(c+2)(a+2),$$
and $3 \le a+b+c$ (using $\frac{9}{a+b+c+6} \le \frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}$). But I didn't know how to deal with the $\sqrt{a}+\sqrt{b}+\sqrt{c}$.
Can anyone help me? Thank you so much.

Complement to @richrow's nice answer
Using $\sqrt{u} \ge \frac{2u}{1 + u}$ for all $u\ge 0$ (easy to prove using AM-GM),
we have
$$\sqrt a = \sqrt 2 \sqrt{a/2}
\ge \sqrt 2 \frac{2 \cdot a/2}{1 + a/2}
= \sqrt 2 \frac{2a}{2 + a} = 2\sqrt 2
- 4\sqrt 2 \frac{1}{2 + a}.$$
Thus, we have
$$\sqrt{a} + \sqrt{b} + \sqrt{c}
\ge 6\sqrt 2 - 4\sqrt 2 \left(\frac{1}{2 + a} + \frac{1}{2 + b} + \frac{1}{2 + c}\right) = 2\sqrt 2$$
where we have used (see OP)
$$\frac{1}{2 + a} + \frac{1}{2 + b} + \frac{1}{2 + c} - 1
= \frac{4 - ab - bc - ca - abc}{(2 + a)(2 + b)(2 + c)} = 0.$$
Also, when $a = b = 2, c = 0$, we have
$ab + bc + ca + abc = 4$ and $\sqrt{a} + \sqrt{b} + \sqrt{c} = 2\sqrt 2$.
Thus, the minimum is $2\sqrt 2$.
