Find minimal value of $\left(2-x\right)\left(2-y\right)\left(2-z\right)$ 
Let $x,y,z>0$ such that $x^2+y^2+z^2=3$. Find minimal value of $$\left(2-x\right)\left(2-y\right)\left(2-z\right)$$

I thought the equality occurs at $x = y = z = 1$ (then it is easy), but the fact is $x = y = \frac{1}{3}; z = \frac{5}{3}$. So I just thought of using $uvw$, but I am not allowed to use it during my exam. Because of the equality I cannot use AMGM, Cauchy-Schwarz, etc.
I tried to use Mixing-Variables, but I failed. Please help.

From Inequality on AoPS:
Without loss of generality $x \ge y \ge z > 0$, so that $z \le 1$. We have
$$
2(2-x)(2-y) = (x+y-2)^2 +4 - x^2-y^2 \ge 4 - x^2 - y^2  = 1+z^2
$$
and therefore
$$
 (2-x)(2-y)(2-z) \ge \frac 1 2 (1+z^2)(2-z) =: f(z) \, .
$$
An elementary calculation shows that the minimum of $f$ on $[0, 1]$ is $f(1/3) = 25/27$, i.e.
$$
(2-x)(2-y)(2-z) \ge \frac{25}{27} \, .
$$
Equality holds if $(x, y, z)$ is a permutation of $(5/3, 1/3, 1/3)$.
