If $\sum_{i=1}^{n}x_{i}=1$ and $x_{i}\ge 0$ then $\sum_{1\le i

Question

Show that $$\sum_{1\le i<j\le n}x_{i}x_{j}(x_{i}+x_{j})\le\dfrac{1}{4}$$

for all $ n$-tuples $ (x_1, \ldots, x_n)$ satisfying $ x_i \geq 0$ and $ \sum_{i=1}^{n} x_i =1.$

I tried C-S, but without success.

Answer 1

We have $$ \sum_{1\le i<j\le n}x_{i}x_{j}(x_{i}+x_{j}) = \sum_{i, j=1}^n x_i^2 x_j - \sum_{i=1}^n x_i^3 = \sum_{i=1}^n x_i^2 - \sum_{i=1}^n x_i^3 $$ because $\sum_{j=1}^n x_j = 1$. It follows that $$ \sum_{1\le i<j\le n}x_{i}x_{j}(x_{i}+x_{j}) = \sum_{i=1}^n x_i^2 (1-x_i) \le \frac 14 \sum_{i=1}^n x_i = \frac 14 $$ because $x(1-x) \le \frac 14$ for $0 \le x \le 1$.

The bound is sharp (for $n \ge 2$) as can be seen by choosing $$ (x_1, x_2, x_3, \ldots, x_n) = (\frac 12, \frac 12, 0, \ldots, 0) \, . $$