14. Let $x_{i} \geqslant 0, i=1,2, \cdots, n$, and $\sum_{i=1}^{n} x_{i}=1, n \geqslant 2$. Find the maximum value of $\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}+x_{j}\right)$. (32nd IMO Shortlist)
14. The maximum value of $\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}+x_{j}\right)$ is $\frac{1}{4}$. The proof is as follows:
$$\begin{array}{l}
x_{1} x_{2}\left(x_{1}+x_{2}\right)+x_{1} x_{3}\left(x_{1}+x_{3}\right)+\cdots+ \\
x_{1} x_{n}\left(x_{1}+x_{n}\right)+x_{2} x_{3}\left(x_{2}+x_{3}\right)+\cdots+ \\
x_{n-1} x_{n}\left(x_{n-1}+x_{n}\right) \leqslant \frac{1}{4}
\end{array}$$

Let the left side of equation (1) be denoted as $A$. Then, noting that $x_{1}+x_{2}+x_{3}+\cdots+x_{n}=1$, we have
$$\begin{aligned}
A= & x_{1}^{2}\left(x_{2}+x_{3}+\cdots+x_{n}\right)+x_{2}^{2}\left(x_{1}+x_{3}+\cdots+x_{n}\right)+\cdots+ \\
& x_{n}^{2}\left(x_{1}+x_{2}+\cdots+x_{n-1}\right)= \\
& x_{1}^{2}\left[\left(x_{1}+x_{2}+x_{3}+\cdots+x_{n}\right)-x_{1}\right]+ \\
& x_{2}^{2}\left[\left(x_{1}+x_{2}+x_{3}+\cdots+x_{n}\right)-x_{2}\right]+\cdots+ \\
& x_{n}^{2}\left[\left(x_{1}+x_{2}+\cdots+x_{n}\right)-x_{n}\right]= \\
& x_{1}^{2}\left(1-x_{1}\right)+x_{2}^{2}\left(1-x_{2}\right)+\cdots+x_{n}^{2}\left(1-x_{n}\right)= \\
& \left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)-\left(x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}\right)
\end{aligned}$$

Since $x_{i} \geqslant 0, i=1,2, \cdots, n$, by the Cauchy-Schwarz inequality, we have
$$\left(x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}\right)\left(x_{1}+x_{2}+\cdots+x_{n}\right) \geqslant\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)^{2}$$

Thus,
$$\begin{aligned}
A \leqslant & \left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)-\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)^{2}= \\
& -\left[\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)-\frac{1}{2}\right]^{2}+\frac{1}{4} \leqslant \frac{1}{4}
\end{aligned}$$

By setting two of the numbers $x_{1}, x_{2}, \cdots, x_{n}$ to $\frac{1}{2}$ and the rest to 0, the equality can be achieved.