44. Given an acute triangle $\triangle ABC$ with side lengths $a, b, c$, and positive real numbers $x, y, z$ satisfying $\frac{a y z}{x}+\frac{b z x}{y}+\frac{c x y}{z}=P$, where $P$ is a given positive real number. Find the maximum value of $S=(b+c-a) x^{2}+(c+a-b) y^{2}+(a+b-c) z^{2}$, and determine the values of $x, y, z$ when $S$ attains this maximum value. (2007 Anhui Province Mathematics Competition)
44. Since $\triangle A B C$ is an acute triangle, we have $a^{2}+b^{2}>c^{2}, b^{2}+c^{2}>a^{2}, c^{2}+a^{2}>b c^{2}$. Therefore, by the mean value inequality, we get
$$\begin{array}{l}
\left(b^{2}+c^{2}-a^{2}\right) x^{2}+\left(c^{2}+a^{2}-b^{2}\right) y^{2}+\left(a^{2}+b^{2}-c^{2}\right) z^{2} \leqslant \\
\frac{1}{2}\left(b^{2}+c^{2}-a^{2}\right) x^{2}\left(\frac{y^{2}}{z^{2}}+\frac{z^{2}}{y^{2}}\right)+\left(c^{2}+a^{2}-b^{2}\right) y^{2}\left(\frac{z^{2}}{x^{2}}+\frac{x^{2}}{z^{2}}\right)+ \\
\left(a^{2}+b^{2}-c^{2}\right) z^{2}\left(\frac{x^{2}}{y^{2}}+\frac{y^{2}}{x^{2}}\right)=\frac{a^{2} y^{2} z^{2}}{x^{2}}+\frac{b^{2} z^{2} x^{2}}{y^{2}}+\frac{c^{2} x^{2} y^{2}}{z^{2}}= \\
\left(\frac{a y z}{x}+\frac{b z x}{y}+\frac{c x y}{z}\right)^{2}-2\left(b c x^{2}+c a y^{2}+a b z^{2}\right)
\end{array}$$

Thus,
$$\begin{array}{l}
{\left[(b+c)^{2}-a^{2}\right] x^{2}+\left[(c+a)^{2}-b^{2}\right] y^{2}+\left[(a+b)^{2}-c^{2}\right] z^{2} \leqslant} \\
\left(\frac{a y z}{x}+\frac{b z x}{y}+\frac{c x y}{z}\right)^{2}=P^{2}
\end{array}$$

That is, $(a+b+c) S \leqslant P^{2}, S \leqslant \frac{P^{2}}{a+b+c}$.
The equality holds if and only if $x^{2}=y^{2}=z^{2}$, i.e., $x=y=z=\frac{P}{a+b+c}$.
Therefore, the maximum value sought is $\frac{P^{2}}{a+b+c}$, and when $S$ takes this maximum value, $x=y=z=\frac{P}{a+b+c}$.