Answer
So the minimum value $=5 \quad \operatorname{at}(\pm \sqrt{5}, 0,0)$
Work Step by Step
We are given a function $z^{2}+y^{2}+x^{2}=f(x, y)$ and the constraint $g(x, y)=$ $-5+x^{2}-y z$
We want to maximize the function $f$ by using Lagrange multipliers,
\[
\lambda \nabla g=\nabla f
\]
We will need the gradient
\[
\begin{aligned}
\nabla f &=2 x \hat{i}+2 y \hat{j}+2 z \hat{k} \\
\nabla g &=2 x \hat{i}-z \hat{j}-y \hat{k} \\
\because \quad \nabla f &=\lambda \nabla g \\
(2 x \hat{i}+2 y \hat{j}+2 z \hat{k}) &=\lambda(2 x \hat{i}-z \hat{j}-y \hat{k})
\end{aligned}
\]
Which is equivalent to the pair of equations:
\[
\begin{aligned}
&\lambda 2 x=2 x \\
&\lambda(-z) = 2y\\
&\lambda(-y) =2z\\
\Rightarrow \frac{2 x}{2 x} &=\frac{2 y}{-z}=\frac{2 z}{-y}
\end{aligned}
\]
If $\lambda \neq\pm 2,$ then $y=z=0;$ thus, from the constraint equation $-5-z y+x^{2}$
\[
5=x^{2}-0 \quad \Rightarrow \pm \sqrt{5}=x
\]
\[
\begin{array}{l}
(\pm \sqrt{5})^{2}+0+0 =f\\
5=f
\end{array}
\]
