19. Suppose $x, y, z$ and $\lambda$ are positive real numbers such that
$$
\begin{aligned}
y z & =6 \lambda x \\
x z & =6 \lambda y \\
x y & =6 \lambda z \\
x^{2}+y^{2}+z^{2} & =1
\end{aligned}
$$

Find the value of $(x y z \lambda)^{-1}$.
19. Answer: 54
Multiplying the first three equations by $x, y$ and $z$ respectively, we have
$$
x y z=6 \lambda x^{2}=6 \lambda y^{2}=6 \lambda z^{2}
$$

Since $\lambda \neq 0$ and $x^{2}+y^{2}+z^{2}=1$, we deduce that $x^{2}=y^{2}=z^{2}=\frac{1}{3}$, so $x=y=z=\frac{1}{\sqrt{3}}$ and $\lambda=\frac{x y z}{6 x^{2}}=\frac{z^{3}}{2 x^{2}}=\frac{1}{6 \sqrt{3}}$.
Hence
$$
(x y z \lambda)^{-1}=(\sqrt{3})^{3} 6 \sqrt{3}=54
$$