Write $f(x_1,x_2,\dots,x_n)=\sum_{i<j}{x_ix_j(x_i+x_j)}$ where the sum runs over all $n$ tuples.Wlog let $x_1 \ge x_2 \ge\dots\ge x_k > x_{k+1}=\dots=x_n=0$We sall now solve the problem by using variable mixing.It is not hard to show that $f(x_1,x_2,\dots,x_{k-1}+x_k,0,0,\dots) > f(x_1,x_2,\dots,x_k0,0,\dots)$.So using this over and over again,we see that it suffices to check the maximum value of $f(x_1,x_2,0,0\dots,0)$ with $x_1+x_2=1$ which is clearly $\frac{1}{4}$,So this is the required answer with equality when $x_1=x_2=\frac{1}{2},x_3=\dots=x_n=0$.

Notice that $\sum_{i < j} x_i x_j (x_i + x_j) = \sum_{i \neq j} x_i^2 x_j = \sum_{i = 1}^n x_i^2 \sum_{j \neq i} x_j$. Using the fact $\sum_{i = 1}^n x_i = 1$, this expression can be rewritten as $\sum_{i = 1}^n x_i^2(1 - x_i)$. Finally, manipulating the inequality $x_i (x_i - 1/2)^2 \geq 0$ yields $x_i^2(1 - x_i) \leq \frac{1}{4} x_i$, thus yielding the result with equality case possible.

Determine the minimum value of the sum \[ \sum_{ i < j} x_ix_j (x_i + x_j) \]over all $ n -$tuples $ (x_1, \ldots, x_n),$ satisfying $ x_i \geq 0$ and $ \sum^n_{i = 1} x_i = 1.$

sqing wrote: Determine the minimum value of the sum \[ \sum_{ i < j} x_ix_j (x_i + x_j) \]over all $ n -$tuples $ (x_1, \ldots, x_n),$ satisfying $ x_i \geq 0$ and $ \sum^n_{i = 1} x_i = 1.$ Let $P$ be $\sum_{i<j}x_ix_j(x_i+x_j)$. Note that $P\ge0$. When $x_1=1$ and $x_2=x_3=\cdots=x_n=0$, $P=0$. Thus the minimum value of $P$ is $0$.