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.
 A: 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 $2\sqrt{2}$?)
A: 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 $2\sqrt 2$.
A: Not beautiful solution.
WLOG $a\leq b\leq c$. Suppose $a>1$, then $b>1$, $c>1$, $ab+bc+ca+abc > 4$. Therefore $a\leq 1$.
Let $\sqrt{b}+\sqrt{c}=d$, $\sqrt{bc}=e$. Then $bc=e^2$, $d^2=b+c+2e$, $b+c=d^2-2e$. $(\sqrt{b}-\sqrt{c})^2=b+c-2e=d^2-4e\geq 0$, therefore $d^2\geq 4e$.
$$ab+bc+ca+abc=4\Rightarrow a(b+c+bc)+bc=4 \Rightarrow a(d^2-2e+e^2)+e^2=4$$
$$d^2\geq 4e \Rightarrow a(4e-2e+e^2)+e^2=\leq 4 \Rightarrow (a+1)e^2+2ae-4\leq 0$$
$$(a+1)e^2+2ae-4=(a+1)e^2+2(a+1)e-2e-4=((a+1)e-2)(e+2)$$
$$(a+1)e^2+2ae-4\leq 0\Rightarrow (a+1)e-2\leq 0 \Rightarrow e\leq\frac2{a+1}$$
$$a(d^2-2e+e^2)+e^2=4\Rightarrow ad^2=4+2ae-(a+1)e^2=4+\frac{a^2}{a+1}-(a+1)\left(e-\frac{a}{a+1}\right)^2$$
$$0\leq e \leq \frac2{a+1}, a\leq 1\Rightarrow \left|e-\frac{a}{a+1}\right|\leq \frac{2-a}{a+1}\Rightarrow ad^2\geq 4+\frac{a^2}{a+1}-(a+1)\left(\frac{2-a}{a+1}\right)^2\Rightarrow$$
$$ad^2\geq \frac{8a}{a+1} \Rightarrow d^2\geq \frac8{a+1}\Rightarrow d\geq\frac{2\sqrt{2}}{\sqrt{a+1}}$$
$$\sqrt{a}+\sqrt{b}+\sqrt{c}=\sqrt{a}+d\geq \sqrt{a}+\frac{2\sqrt{2}}{\sqrt{a+1}}$$
Let $f=\sqrt{a}+\frac{2\sqrt{2}}{\sqrt{a+1}}$. Then $\sqrt{a}+\sqrt{b}+\sqrt{c}\geq f$.
$$f^2=a+\frac{8}{a+1}+\frac{4\sqrt{2a}}{\sqrt{a+1}}=a+8-\frac{8a}{a+1}+\frac{4\sqrt{2a}}{\sqrt{a+1}}-1+1=8+a+1-\left(2\sqrt{\frac{2a}{a+1}}-1\right)^2$$
$$a\leq 1 \Rightarrow a+a\leq a+1\Rightarrow 2\sqrt{\frac{2a}{a+1}}\leq 2\Rightarrow \left|2\sqrt{\frac{2a}{a+1}}-1\right|\leq 1\Rightarrow$$
$$f^2\geq 8+a+1-1=8+a\Rightarrow \sqrt{a}+\sqrt{b}+\sqrt{c}\geq f\geq \sqrt{8+a}\geq\sqrt{8}=2\sqrt{2}$$
Minimum is obtained at $a=0$ and $b=c$, because it requires $8+a=8$ and $\left|e-\frac{a}{a+1}\right|=\frac{2-a}{a+1}\Rightarrow e=\frac{2}{a+1}\Rightarrow d^2=4e\Rightarrow (\sqrt{b}-\sqrt{c})^2=0$.
