First solution. Using the given condition, the inequality can be written equivalently $\sqrt{xy} + \sqrt{yz} + \sqrt{zx} \ge \sqrt{3(2xy + 2yz + 2zx - x^2 - y^2 - z^2)}$ or, denoting $\sqrt{x} = a, \sqrt{y} = b, \sqrt{z} = c$, we have $(ab + bc + ca)^2 \ge 3(2a^2b^2 + 2b^2c^2 + 2c^2a^2 - a^4 - b^4 - c^4)$. We thus have to prove that $3(a^4 + b^4 + c^4) + 2(a^2bc + b^2ca + c^2ab) \ge 5(a^2b^2 + b^2c^2 + c^2a^2)$.
The following inequality is well known (Schur): $a^2(a-b)(a-c) + b^2(b-c)(b-a) + c^2(c-a)(c-b) \ge 0$.
Multiplied by 2, it becomes $2(a^4 + b^4 + c^4) + 2(a^2bc + b^2ca + c^2ab) \ge 2(a^3b + a^3c + b^3a + b^3c + c^3a + c^3b)$.
But $2(a^3b + a^3c + b^3a + b^3c + c^3a + c^3b) = 2ab(a^2 + b^2) + 2bc(b^2 + c^2) + 2ca(c^2 + a^2) \ge 4a^2b^2 + 4b^2c^2 + 4c^2a^2$ and $a^4 + b^4 + c^4 \ge a^2b^2 + b^2c^2 + c^2a^2$. Adding these three inequalities given the desired one.

Second solution. Rewrite $x^2 + y^2 + z^2 + 3 = 2(xy + yz + zx)$ as
$$(\sqrt{x} + \sqrt{y} + \sqrt{z})(\sqrt{x} + \sqrt{y} - \sqrt{z})(\sqrt{x} + \sqrt{z} - \sqrt{y})(\sqrt{y} + \sqrt{z} - \sqrt{x}) = 3.$$
$$\text{Let } \sqrt{x} + \sqrt{y} - \sqrt{z} = 2a, \sqrt{z} + \sqrt{y} - \sqrt{x} = 2b, \sqrt{x} + \sqrt{z} - \sqrt{y} = 2c.$$
Then
$$abc(a + b + c) = \frac{3}{16}. \qquad (1)$$
The sum $a + b + c = \frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{2}$ is positive. This means that either $a, b, c$ are all positive, or exactly two of them are negative (according to (1)). If, say, $a$ and $b$ are negative, then so is their sum, i.e. $a + b = \sqrt{x} < 0$, which is false.
All that remains to be proven is
$$(a + b)(a + c) + (b + a)(b + c) + (c + a)(c + a) \ge 3.$$
But, $\sum_{cyc} (a+b)(a+c) = \sum_{cyc} a^2 + 3 \sum_{cyc} ab \geq 4 \sum_{cyc} ab \geq 4\sqrt{3abc(a+b+c)} = 3$,
which is exactly what we wanted. Equality holds if and only if $x = y = z = 1$.