If $xy+xz+yz=1+2xyz$ then $\sqrt{x}+\sqrt{y}+\sqrt{z}\geq2$.

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 $

$$p=a+b+c\ , \ q=ab+bc+ca \ , \ r=abc$$

$$q^2=1+2pr+2r^2 \Rightarrow q \ge 1$$

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

2. $1 \le q \le \dfrac{4}{3}\ , \ 2pr = q^2-1-2r^2 \le q^2-1$

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 : $a=b=1 \ , \ c=0$

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.