longlong123 wrote: Let $n$ be a positive integer. The real numbers $a_1,a_2,\cdots,a_n$ and $r_1,r_2,\cdots,r_n$ are such that $a_1\leq a_2\leq \cdots \leq a_n$ and $0\leq r_1\leq r_2\leq \cdots \leq r_n$. Prove that $\sum_{i=1}^n\sum_{j=1}^n a_i a_j \min (r_i,r_j)\geq 0$ It is here. The ordering is not needed, only $r_i\ge 0$ .

For $\forall 1\leq i\leq n,$ define $f_i(x)=\begin{cases}1& 0\leq x\leq r_i\\ 0 & x>r_i\end{cases}.$ Then $$\begin{aligned}\sum_{i=1}^n\sum_{j=1}^n a_i a_j \min \{r_i,r_j\}&=\sum_{i=1}^n\sum_{j=1}^n a_i a_j \int_{0}^{+\infty}f_i(x)f_j(x)dx\\ &=\int_{0}^{+\infty}\sum_{i=1}^n\sum_{j=1}^n a_i a_jf_i(x)f_j(x)dx\\ &=\int_{0}^{+\infty}\left(\sum_{i=1}^na_if_i(x)\right)^2dx\\&\geqslant 0.\blacksquare\end{aligned}$$

One of my fav trick