10. (20 points) Let $x_{1}, x_{2}, x_{3}$ be non-negative real numbers, satisfying $x_{1}+x_{2}+x_{3}=1$. Find the minimum and maximum values of
$$
\left(x_{1}+3 x_{2}+5 x_{3}\right)\left(x_{1}+\frac{x_{2}}{3}+\frac{x_{3}}{5}\right)
$$
10. By Cauchy's inequality, we have
$$
\begin{array}{l}
\left(x_{1}+3 x_{2}+5 x_{3}\right)\left(x_{1}+\frac{x_{2}}{3}+\frac{x_{3}}{5}\right) \\
\geqslant\left(\sqrt{x_{1}} \sqrt{x_{1}}+\sqrt{3 x_{2}} \sqrt{\frac{x_{2}}{3}}+\sqrt{5 x_{3}} \sqrt{\frac{x_{3}}{5}}\right)^{2} \\
=\left(x_{1}+x_{2}+x_{3}\right)^{2}=1,
\end{array}
$$

When $x_{1}=1, x_{2}=0, x_{3}=0$, the equality holds, so the desired minimum value is 1.
$$
\begin{array}{l}
\text { Also, }\left(x_{1}+3 x_{2}+5 x_{3}\right)\left(x_{1}+\frac{x_{2}}{3}+\frac{x_{3}}{5}\right) \\
=\frac{1}{5}\left(x_{1}+3 x_{2}+5 x_{3}\right)\left(5 x_{1}+\frac{5 x_{2}}{3}+x_{3}\right) \\
\leqslant \frac{1}{5} \times \frac{1}{4}\left(\left(x_{1}+3 x_{2}+5 x_{3}\right)+\left(5 x_{1}+\frac{5 x_{2}}{3}+x_{3}\right)\right)^{2} \\
=\frac{1}{20}\left(6 x_{1}+\frac{14}{3} x_{2}+6 x_{3}\right)^{2} \\
\leqslant \frac{1}{20}\left(6 x_{1}+6 x_{2}+6 x_{3}\right)^{2}=\frac{9}{5},
\end{array}
$$

When $x_{1}=\frac{1}{2}, x_{2}=0, x_{3}=\frac{1}{2}$, the equality holds, so the desired maximum value is $\frac{9}{5}$.