More Algebra

$x^{2} + y^{2} + z^{2} = 4$ describes a sphere of radius $2$ centred at the origin. $(x - 1)^{2} + (y + 2)^{2} + (z - 2)^{2} = r^{2}$ describes a sphere of radius $r$ centred at $(1, -2, 2)$ , so the maximum value of $r$ given that $x^{2} + y^{2} + z^{2} = 4$ will be the greatest distance from $(1, -2, 2)$ to a point on the radius $2$ sphere. Since $(1, -2, 2)$ is a distance of $\sqrt{1^{2} + (-2)^{2} + 2^{2}} = 3$ from the origin, the greatest distance, and hence greatest value for $r$ , will be $3$ plus the radius of the given circle, i.e., $3 + 2 = 5$ , giving a maximum for $r^{2}$ of $\boxed{25}$ .

First let us consider expanding the expression $(x-1)^2+(y+2)^2+(z-2)^2=x^2+y^2+z^2-2x+4y-4z+9.$ We then can substitute $x^2+y^2+z^2=4$ into this to get $-2x+4y-4z+13.$ We need to maxmize this, and one method of doing it is by using the Cauchy-Swartz inequality of the dot product. To do this, let us write the expression as the dot product of two vectors. We want the maximum, so it only makes since to make sure we are getting a positive dot product: $-2x+4y-4z+13=\left| \left[ \begin{matrix} -2 \\ 4 \\ -4 \end{matrix} \right] \cdot \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \right|+13$ By the Cauchy-Swartz inequality: $\left| \left[ \begin{matrix} -2 \\ 4 \\ -4 \end{matrix} \right] \cdot \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \right|+13\le \left| \left[ \begin{matrix} -2 \\ 4 \\ -4 \end{matrix} \right] \right| \left| \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \right|+13=\sqrt{(-2)^2+(4)^2+(-4)^2}\cdot \sqrt{4}+13$ $=\boxed{25}$

Let $f(x,y,z)$ be the expression. Then

$\begin{aligned} f(x,y,z) & = (x-1)^2 + (y+2)^2 + (z-2)^2 \\ & = \blue{x^2} - 2x + 1 + \blue{y^2} + 4y + 4 + \blue{z^2} - 4z + 4 & \small \blue{\text{Since }x^2+y^2+z^2 = 4} \\ & = 13 - 2x + 4y - 4z \end{aligned}$

For maximum $f(x,y,z)$ , $-2x > 0$ and $-4z > 0$ or $x, z < 0$ . Let $a = - x$ , $b=y$ , and $c=-z$ . Then $a^2+b^2+c^2 = x^2+y^2+z^2 = 4$ and $f(x,y,z)$ = 13+2a+4b+4c). By Cauchy-Schwarz inequality

$\begin{aligned} 13+2a+4b+4c \le 13+\sqrt{(2^2+4^2+4^2)(a^2+b^2+c^2)} = 13 + \sqrt{(36)(4)} = 13+12 = \boxed{25} \end{aligned}$

Equality occurs when $a=\frac 23, b = \frac 43, c = \frac 43$ or $x=-\frac 23, y = \frac 43, z = - \frac 43$ .