15. A1 (POL) $)^{1 \mathrm{MO} 2}$ Let $n \geq 2$ be a fixed integer. Find the least constant $C$ such that the inequality
$$
\sum_{i<j} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right) \leq C\left(\sum_{i} x_{i}\right)^{4}
$$
holds for every $x_{1}, \ldots, x_{n} \geq 0$ (the sum on the left consists of $\binom{n}{2}$ summands). For this constant $\bar{C}$, characterize the instances of equality.
15. For all $x_{i}=0$ any $C$ will do, so we may assume the contrary. Since the equation is symmetric and homogeneous, we may assume $\sum_{i} x_{i}=1$. The equation now becomes $F\left(x_{1}, x_{2}, \ldots, x_{n}\right)=\sum_{i<j} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right)=$ $\sum_{i} x_{i}^{2} \sum_{j \neq i} x_{j}=\sum_{i} x_{i}^{3}\left(1-x_{i}\right)=\sum_{i} f\left(x_{i}\right) \leq C$, where we define $f(x)=$ $x^{3}-x^{4}$. We note that for $x, y \geq 0$ and $x+y \leq 2 / 3$,
$$
f(x+y)+f(0)-f(x)-f(y)=3 x y(x+y)\left(\frac{2}{3}-x-y\right) \geq 0 .
$$

We note that if at least three elements of $\left\{x_{1}, x_{2}, \ldots, x_{n}\right\}$ are nonzero the condition of (1) always holds for the two smallest ones. Hence, applying (1) repeatedly, we obtain $F\left(x_{1}, x_{2}, \ldots, x_{n}\right) \leq F(a, 1-a, 0, \ldots, 0)=\frac{1}{2}(2 a(1-$ a)) $(1-2 a(1-a)) \leq \frac{1}{8}=F\left(\frac{1}{2}, \frac{1}{2}, 0, \ldots, 0\right)$. Thus we have $C=\frac{2}{8}$ (for all $n$ ), and equality holds only when two $x_{i}$ are equal and the remaining ones are 0 .
Second solution. Let $M=x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}$. Using $a b \leq(a+2 b)^{2} / 8$ we have
$$
\begin{aligned}
\sum_{1 \leq i<j \leq n} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right) & \leq M \sum_{i<j} x_{i} x_{j} \\
& \leq \frac{1}{8}\left(M+2 \sum_{i<j} x_{i} x_{j}\right)^{2}=\frac{1}{8}\left(\sum_{i=1}^{n} x_{i}\right)^{4} .
\end{aligned}
$$

Equality holds if and only if $M=2 \sum_{i<j} x_{1} x_{j}$ and $x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right)=M x_{i} x_{j}$ for all $i<j$, which holds if and only if $n-2$ of the $x_{i}$ are zero and the remaining two are equal.
Remark. Problems (SL90-26) and (SL91-27) are very similar.