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.

Wlog $z\le 1$ because the variables cannot all be $\ge1$.
For fixed $z$, we want to minimize $(2-x)(2-y)$ under a constraint $x^2+y^2=3-z^2$, which is a constant between $2$ and $3$.
Now
$$ \begin{align}(2-x)(2-y)&=4-2(x+y)+xy\\&=4-2(x+y)+\frac12(x+y)^2-\frac{3-z^2}2\\
&=\frac12\left(x+y-2\right)^2+2-\frac{3-z^2}2
\\ &\ge2-\frac{3-z^2}2\\&=\frac{z^2+1}2
\end{align}$$
with equality iff $x+y=2$. Note that this makes $x^2+y^2=x^2+(2-x)^2=4-4x+2x^2=2(x-1)^2+2$, i.e., we can always find such $x,y$ when $z\le 1$ and $x^2+y^2=3-z^2\in [2,3]$.
So we want to minimize $\frac{z^2+1}2\cdot(2-z)=\frac12(2-z+2z^2-z^3)$ with $0<z\le1$.
The derivative of this cubic is $-3z^2+4z-1=(z-1)(1-3z)$, so we find the desired minimum at $z=\frac13$. 
