15. Let real numbers $a, b$ satisfy $a=x_{1}+x_{2}+x_{3}=$ $x_{1} x_{2} x_{3}, a b=x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}$, where $x_{1}, x_{2}, x_{3}>0$. Find the maximum value of $P=\frac{a^{2}+6 b+1}{a^{2}+a}$.
So, $a \geqslant 3 \sqrt{3}$.
Also, $3\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}\right) \leqslant\left(x_{1}+x_{2}+x_{3}\right)^{2}$, then
$3 a b \leqslant a^{2}$, which means $3 b \leqslant a$.
Thus, $P=\frac{a^{2}+6 b+1}{a^{2}+a} \leqslant \frac{a^{2}+2 a+1}{a^{2}+a}$
$$
=1+\frac{1}{a} \leqslant 1+\frac{1}{3 \sqrt{3}}=\frac{9+\sqrt{3}}{9} \text {. }
$$

When $x_{1}=x_{2}=x_{3}, a=3 \sqrt{3}, 3 b=a$, i.e., $x_{1}=x_{2}=x_{3}=\sqrt{3}, a=3 \sqrt{3}, b=\sqrt{3}$, the equality in (1) holds.
Therefore, the maximum value of $P$ is $\frac{9+\sqrt{3}}{9}$.