Chebyshev's sum inequality

Expansion.
$$\sum_{i,j=1}^n(a_i-a_j)(b_i-b_j)=\sum_{i,j}a_ib_i+\sum_{i,j}a_jb_j-\sum_{i,j}a_ib_j-\sum_{i,j}a_jb_i.$$
The first two terms each equal $n\sum a_ib_i$ (renaming $j$ to $i$). The last two each equal $(\sum a_i)(\sum b_j)$. So
$$\sum_{i,j}(a_i-a_j)(b_i-b_j)=2n\sum a_ib_i-2\left(\sum a_i\right)\left(\sum b_j\right).$$

Non-negativity. Since $a_i\le a_j\iff b_i\le b_j$ (both increasing), $(a_i-a_j)(b_i-b_j)\ge0$ for all $i,j$. So the sum is $\ge0$.

Conclusion.
$$n\sum a_ib_i\ge\left(\sum a_i\right)\left(\sum b_j\right),$$
which is exactly the claimed inequality.