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.

Here is a sketch of the solution (you need to fill in the details).
One might use the following substitution
$$
a=\frac{2x}{y+z},~b=\frac{2y}{z+x},~c=\frac{2z}{x+y},
$$
where $a,b,c>0$ (why do they exist?). Thus, it remains to find the minimum (or rather infimum) of
$$
F(x,y,z)=\sqrt{\frac{2x}{y+z}}+\sqrt{\frac{2y}{z+x}}+\sqrt{\frac{2z}{x+y}}
$$
for positive $x,y,z$.
If $x=y=t>0$ and $z=0$, then $F(x,y,z)=F(t,t,0)=2\sqrt{2}$, so the desired infimum is at least $2\sqrt{2}$ (note that $z$ is not positive, but the conclusion still holds -- why?).
Finally, to prove that the expression above is always greater than $2\sqrt{2}$ it suffices to observe that
$$
\sqrt{\frac{x}{y+z}}\geq\frac{2x}{x+y+z}.
$$
(Why the last inequality holds? Also, why $F(x,y,z)$ will be strictly greater than $2\sqrt{2}$?)
