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.
 A: 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 $z=\frac13$. 
A: 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 $(5/3, 1/3, 1/3)$.
A: 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 $x, y, z > 0$.
