Prove that if $x,y,z$ are positive real numbers and $ xy+xz+yz = 1$ then $\sqrt{x}+\sqrt{y}+\sqrt{z} > 2$ 
Prove that if $x,y,z$ are positive real numbers and $ xy+xz+yz = 1$ then  $$\sqrt{x}+\sqrt{y}+\sqrt{z} > 2$$

I am having a hard time relating the square roots in the inequality to the given condition. I was thinking that maybe there is some substitution I could do, but I couldn't think of any.

Another simple solution.
We show that
$$(\sqrt{x}+\sqrt{y}+\sqrt{z})^4 \ge 16(xy+yz+zx) \quad\forall x,y,z \ge 0$$
or 
$$(a+b+c)^4 \ge 16(a^2b^2+b^2c^2+c^2a^2) \quad\forall a,b,c \ge 0.$$
WLOG, assume that $a=\max(a,b,c)$, then we have $$a^2b^2+b^2c^2+c^2a^2 = a^2(b+c)^2 + bc(bc-2a^2) \le a^2(b+c)^2.$$
Thus it reduces to prove $$(a+b+c)^4 \ge 16 a^2(b+c)^2$$
or equivalently
$$(a+b+c)^2 \ge 4 a(b+c),$$
which is true because $$(a+b+c)^2 - 4 a(b+c) = (a-b-c)^2 \ge 0.$$
Equality occurs when $a=b$ and $c=0$ or when $a=c$ and $b=0$.
