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. 

Short proof.
Clearly $xy+yz+zx \ge 1$
We have by AM-GM
$$(\sqrt{x}+\sqrt{y}+\sqrt{z})^2=x+y+z+2(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})$$
$$\ge x+y+z+ \frac{4xy}{x+y}+\frac{4yz}{y+z}+\frac{4zx}{z+x} $$
$$ \ge x+y+z + \frac{4(xy+yz+zx)}{x+y+z} \ge x+y+z+\frac{4}{x+y+z} \ge 4$$
The proof is complete.
