16．记$\triangle ABC$的内角$A$，$B$，$C$的对边分别为$a$，$b$，$c$，已知$\cos B = \frac{2}{3}$，$P$为$\triangle ABC$内一点，若点$P$满足$\overrightarrow{CP} + \frac{b}{a}\overrightarrow{BP} = \frac{a}{c}\overrightarrow{PA}$，且$\overrightarrow{BP} = x\overrightarrow{BA} + y\overrightarrow{BC}$，则$x + y$的最大值为．
解 依题可知$\overrightarrow{BP} = \frac{a}{b}\overrightarrow{PA} + \frac{c}{b}\overrightarrow{CP}$
$\therefore \overrightarrow{BP} = \frac{a}{b}(\overrightarrow{BA} - \overrightarrow{BP}) + \frac{c}{b}(\overrightarrow{BC} - \overrightarrow{BP})$
$\therefore \frac{a + b + c}{b}\overrightarrow{BP} = \frac{a}{b}\overrightarrow{BA} + \frac{c}{b}\overrightarrow{BC}$
$\therefore \overrightarrow{BP} = \frac{a}{a + b + c}\overrightarrow{BA} + \frac{c}{a + b + c}\overrightarrow{BC}$
$\because \overrightarrow{BP} = x\overrightarrow{BA} + y\overrightarrow{BC}$
$\therefore x = \frac{a}{a + b + c}$，$y = \frac{c}{a + b + c}$
$\therefore x + y = \frac{a + c}{a + b + c} = 1 - \frac{b}{a + b + c}$
$\because b^2 = a^2 + c^2 - 2ac\cos B$，$\cos B = \frac{2}{3}$
$\therefore (a + c)^2 - b^2 = \frac{10}{3}ac \leq \frac{5}{6}(a + c)^2$
$\therefore a + c \leq \sqrt{6}b$
$\therefore x + y = 1 - \frac{b}{a + b + c}$
$= 1 - \frac{1}{\frac{a + c}{b} + 1} \leq \frac{6 - \sqrt{6}}{5}$