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.

There is also a straight-forward solution with Lagrange multipliers. It is more convenient to examine the (continuous) function
$$
 f(x, y, z) = (2-x)(2-y)(2-z)
$$
on the (compact) set $K = \{ (x, y, z) \in \Bbb R^3 \mid x^2+y^2+z^2 = 3 \}$ first, without the restriction to positive arguments.
$f$ attains both minimum and maximum on $K$, and the Lagrange method gives that at the extremal points
$$
 -(2-y)(2-z) = \lambda 2x \\
 -(2-x)(2-z) = \lambda 2y \\
 -(2-x)(2-y) = \lambda 2z 
$$
for some $\lambda \in \Bbb R$. We see that $\lambda$ cannot be zero, so that
$$
 x(2-x) = y(2-y) = z(2-z) =: \alpha \, ,
$$
i.e. $x, y, z$ are all solutions of the quadratic equation $t^2 - 2t + \alpha = 0$. It follows that any two of them are equal or add to $2$. So we have two possible cases: 


*

*$x=y=z$. Then necessarily $x=y=z = \pm 1$.

*Up to a permutation, $x=y=2-z$. Substituting that into $x^2+y^2+z^2$ gives
$$
 3 = 2x^2 + (2-x)^2 = 3x^2 -4x + 4 \iff (3x-1)(x-1) = 0 \, .
$$
If $x=1$ then (again) $x=y=z=\pm 1$. Otherwise $(x, y, z) = (1/3, 1/3, 5/3)$.
So the only possible extremal values of $f$ on $K$ are
$$
\begin{align}
 f(1, 1, 1) &= 1 \, \\
 f(-1, -1, -1) &= 27 \, \\
 f(1/3, 1/3, 5/3) &= 25/27 \, ,
\end{align}
$$
and we conclude that $25/27$ is the minimum.
This is also the mimimum for the given problem because it is attained at a point with $x, y, z > 0$.
