If $xy+xz+yz=1+2xyz$ then $\sqrt{x}+\sqrt{y}+\sqrt{z}\geq2$. Let $x$, $y$ and $z$ be non-negative numbers such that $xy+xz+yz=1+2xyz$. Prove that:
$$\sqrt{x}+\sqrt{y}+\sqrt{z}\geq2$$
The equality occurs for $x=y=1$ and $z=0$.
I tried Lagrange Multipliers and more, but I don't see a proof. 

Let: $x=a^2 , y=b^2 , z=c^2$
So we must prove : $a+b+c \ge 2$
for nonnegative $a, b, c   :    \ a^2b^2+b^2c^2+c^2a^2=1+2a^2b^2c^2 $
$$p=a+b+c\ , \ q=ab+bc+ca \ , \ r=abc$$
$$q^2=1+2pr+2r^2 \Rightarrow q \ge 1$$


*

*$q \ge \dfrac{4}{3} \Rightarrow p^2 \ge 3q \ge 4$

*$1 \le q \le \dfrac{4}{3}\  , \ 2pr = q^2-1-2r^2 \le q^2-1$ 
By Shur we have : $p^3+9r \ge 4pq \Rightarrow p^4-4qp^2+\dfrac{9}{2}(q^2-1)\ge 0\Rightarrow$
$$p^2\ge 2q+\sqrt{\dfrac{9-q^2}{2}} \ge 4 \Leftrightarrow (q-1)(23-9q)\ge 0$$
Equality holdes for : $a=b=1 \ , \ c=0$
