9. In an acute $\triangle A B C$, find a point $P$ such that $(B L)^{2}+(C M)^{2}+(A N)^{2}$ is minimized, where $L, M, N$ are the feet of the perpendiculars from $P$ to $B C, C A, A B$ respectively.
Just let $S_{\triangle A^{\prime} B^{\prime} C}$ be simplified to
$S_{\triangle A^{\prime} B^{\prime} C^{\prime}}=2 R^{2} \cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2}$, then to prove $S_{\triangle A^{\prime} B^{\prime} C} \geqslant S_{\triangle A B C}$, it is equivalent to proving the inequality $\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \leqslant \frac{1}{8}$ holds, and this inequality is known to be true.
9. As shown in Figure 21-14, we have $x^{2}+h_{1}^{2}=h_{3}^{2}+(c-z)^{2}, y^{2}$ $+h_{2}^{2}=h_{1}^{2}+(a-x)^{2}, z^{2}+h_{3}^{2}=h_{2}^{2}+(b-y)^{2}$.

Thus $x^{2}+y^{2}+z^{2}=(a-x)^{2}+(b-y)^{2}+(c$. $-z)^{2}$.
$$
\begin{array}{l} 
2\left(x^{2}+y^{2}+z^{2}\right) \geqslant \frac{1}{2}(x+a-x)^{2}+\frac{1}{2}(y+b \\
-y)^{2}+\frac{1}{2}(z+c-z)^{2} \\
= \frac{1}{2}\left(a^{2}+b^{2}+c^{2}\right),
\end{array}
$$

which means $x^{2}+y^{2}+z^{2} \geqslant \frac{1}{4}\left(a^{2}+b^{2}+c^{2}\right)$, with equality holding if and only if $P$ is the circumcenter.