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.
 A: Assume w.l.o.g. that $z \leq x,y$.
We have $(x+z)(y+z) =1+z^2$ and hence $\sqrt{(x+z)}+\sqrt{(y+z)} \geq 2(1+z^2)^{1/4}>2$.
Now it is sufficient to prove: $\sqrt{x}+\sqrt{z}/2 \geq \sqrt{x+z}$ and $\sqrt{y}+\sqrt{z}/2 \geq \sqrt{y+z}$.
It is sufficient to prove the first inequality which is equivalent to:
$ x+z/4+\sqrt{xz} \geq x+z$, or $\sqrt{x} \geq 3\sqrt{z}/4$ which is true.
A: 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 $b=0$.
A: My attempt (not a full solution). The best I was able to achieve is this inequality:

$$\sqrt{x}+\sqrt{y}+\sqrt{z}>\sqrt{2+\sqrt{3}}=1.93185\dots$$

We will use the mean inequalities:
$$\sqrt{\frac{x^2+y^2+z^2}{3}} \geq \frac{x+y+z}{3} \geq \sqrt[3]{xyz} \geq \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$$
By squaring the first inequality and using $xy+xz+yz=1$ and $x,y,z>0$ we obtain:
$$x^2+y^2+z^2 \geq 1$$
By adding $2=2(xy+xz+yz)$ we get:
$$x+y+z \geq \sqrt{3}$$

Additionally from the GM-HM inequality and using $xy+xz+yz=1$ we obtain:
$$\sqrt[3]{xyz} \geq 3 xyz$$
$$xyz \leq \frac{1}{\sqrt{27}}=\frac{1}{3\sqrt{3}}$$
I don't know how to use it here though.

Now let's denote:
$$a=\sqrt{x}+\sqrt{y}+\sqrt{z}$$
$$a^2=x+y+z+2(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}) \geq \sqrt{3}+2(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})$$
$$(a^2-\sqrt{3})^2  \geq 4 (1+2(z \sqrt{xy}+x \sqrt{yz}+y \sqrt{zx}))$$
Notice that:
$$z \sqrt{xy}+x \sqrt{yz}+y \sqrt{zx}=a \sqrt{xyz}$$
Then we obtain a polynomial inequality:
$$(a^2-\sqrt{3})^2  \geq 4 (1+2a \sqrt{xyz})$$
We need a lower bound for $xyz$, but the best I could come up with:
$$xyz>0$$
Using this trivial bound we obtain:
$$(a^2-\sqrt{3})^2  > 4$$
The positive solution is:
$$a>\sqrt{2+\sqrt{3}}$$

I'll update if I manage to solve the original problem.

This may be useful (I got it by applying GM-HM inequality to three pairs of numbers):
$$\sqrt{xy}+\sqrt{yz}+\sqrt{zx} \geq 2\frac{1+(x+y+z)xyz}{x+y+z-xyz}$$
And also by AM-GM for $\sqrt{xy},\sqrt{yz},\sqrt{zx}$ we have:
$$\sqrt{xy}+\sqrt{yz}+\sqrt{zx} \geq 3 \sqrt[3]{xyz}$$
