10. 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 $\left(x_{1}+3 x_{2}+5 x_{3}\right)\left(x_{1}+\frac{x_{2}}{3}+\frac{x_{3}}{5}\right)$.
Solve:
\[
\left(x_{1}+3 x_{2}+5 x_{3}\right)\left(x_{1}+\frac{x_{2}}{3}+\frac{x_{3}}{5}\right) \geqslant\left(\sqrt{x_{1}} \cdot \sqrt{x_{1}}+\sqrt{3 x_{2}} \cdot \sqrt{\frac{x_{2}}{3}}+\sqrt{5 x_{3}} \cdot \sqrt{\frac{x_{3}}{5}}\right)^{2}
\]
\[
=\left(x_{1}+x_{2}+x_{3}\right)^{2}=1, \text{ equality holds when } \left\{\begin{array}{l}\sqrt{x_{1}}=\lambda \sqrt{x_{1}}, \\ \sqrt{\frac{x_{2}}{3}}=\lambda \sqrt{3 x_{2}}, \\ \sqrt{\frac{x_{3}}{5}}=\lambda \sqrt{5 x_{3}} .\end{array}\right.
\]
Thus, when one of \(x_{1}, x_{2}, x_{3}\) is 1 and the other two are 0, \(\left(x_{1}+3 x_{2}+5 x_{3}\right)\left(x_{1}+\frac{x_{2}}{3}+\frac{x_{3}}{5}\right)\) achieves its minimum value of 1. On the other hand, since
\[
\begin{array}{l}
x_{1}+x_{2}+x_{3}=1 \Rightarrow x_{3}=1-x_{1}-x_{2} \Rightarrow\left(x_{1}+3 x_{2}+5 x_{3}\right)\left(x_{1}+\frac{x_{2}}{3}+\frac{x_{3}}{5}\right) \\
=\left[x_{1}+3 x_{2}+5\left(1-x_{1}-x_{2}\right)\right]\left[x_{1}+\frac{x_{2}}{3}+\frac{1}{5}\left(1-x_{1}-x_{2}\right)\right] \\
=-\frac{16}{5} x_{1}^{2}-\frac{32}{15} x_{1} x_{2}-\frac{4}{15} x_{2}^{2}+\frac{16}{5} x_{1}+\frac{4}{15} x_{2}+1 \\
=-\frac{16}{5}\left[x_{1}^{2}+\left(\frac{2}{3} x_{2}-1\right) x_{1}\right]-\frac{4}{15} x_{2}^{2}+\frac{4}{15} x_{2}+1 \\
=-\frac{16}{5}\left(x_{1}+\frac{1}{3} x_{2}-\frac{1}{2}\right)^{2}+\frac{16}{5}\left(\frac{1}{3} x_{2}-\frac{1}{2}\right)^{2}-\frac{4}{15} x_{2}^{2}+\frac{4}{15} x_{2}+1 \\
=-\frac{16}{5}\left(x_{1}+\frac{1}{3} x_{2}-\frac{1}{2}\right)^{2}+\frac{4}{45} x_{2}^{2}-\frac{4}{5} x_{2}+\frac{9}{5} \\
=-\frac{16}{5}\left(x_{1}+\frac{1}{3} x_{2}-\frac{1}{2}\right)^{2}+\frac{4}{45} x_{2}\left(x_{2}-9\right)+\frac{9}{5} \leqslant \frac{9}{5} \text {, } \\
\end{array}
\]
equality holds when \(\left\{\begin{array}{l}x_{1}+\frac{1}{3} x_{2}-\frac{1}{2}=0, \\ x_{2}=0\end{array} \Rightarrow\left\{\begin{array}{l}x_{1}=x_{3}=\frac{1}{2}, \\ x_{2}=0 .\end{array}\right.\right.\)
Therefore, the maximum value of \(\left(x_{1}+3 x_{2}+5 x_{3}\right)\left(x_{1}+\frac{x_{2}}{3}+\frac{x_{3}}{5}\right)\) is \(\frac{9}{5}\).