China Western Mathematical Olympiad

As $x_1,x_2,\dots ,x_n\in [0,1]$, $x_i{x}_j\le x_i$, so we have $$3\sum _{1\le k<l\le n}k{x}_k{x}_l=\sum _{1\le k<l\le n}3k{x}_k{x}_l\le \sum _{1\le k<l\le n}(k{x}_k+2k{x}_l).$$ For $1\le k\le n$, the coefficient of $x_k$ in the last sum is $$2[1+2+\cdots +(k-1)]+k(n-k)=k(n-1),$$ so we have $$\begin{array}{rl}3\sum _{1\le k<l\le n}k{x}_k{x}_l& \le \sum _{1\le k<l\le n}(k{x}_k+2k{x}_l)=\sum _{k=1}^{n}k(n-1)x_k\\ & =(n-1)\sum _{k=1}^{n}k{x}_k,\end{array}$$ and hence the desired inequality holds. $\square$