8. Proof
$$\text { Let } \begin{aligned} D & =\left(\sum_{k=1}^{n} a_{k}\right)\left(\sum_{k=1}^{n} b_{k}\right)-\left(\sum_{k=1}^{n}\left(a_{k}+b_{k}\right)\right)\left(\sum_{k=1}^{n} \frac{a_{k} b_{k}}{a_{k}+b_{k}}\right) \\ & =\left(\sum_{k=1}^{n} a_{k}\right)\left(\sum_{j=1}^{n} b_{j}\right)-\left(\sum_{k=1}^{n}\left(a_{k}+b_{k}\right)\right)\left(\sum_{j=1}^{n} \frac{a_{j} b_{j}}{a_{j}+b_{j}}\right) \\ & =\sum_{k=1}^{n} \sum_{j=1}^{n} \frac{a_{k} b_{j}\left(a_{j}+b_{j}\right)-a_{j} b_{j}\left(a_{k}+b_{k}\right)}{a_{j}+b_{j}} \end{aligned}$$

By swapping the indices $k, j$, the value of $D$ remains unchanged: $D=\sum_{k=1}^{n} \sum_{j=1}^{n} \frac{a_{j} b_{k}\left(a_{k}+b_{k}\right)-a_{k} b_{k}\left(a_{j}+b_{j}\right)}{a_{k}+b_{k}}$,
Adding (1) and (2), we get $2 D=\sum_{k=1}^{n} \sum_{j=1}^{n} \frac{\left(a_{k} b_{j}-a_{j} b_{k}\right)^{2}}{\left(a_{k}+b_{k}\right)\left(a_{j}+b_{j}\right)} \geqslant 0, D=0$ if and only if $a_{k} b_{j}=a, b_{k}(k, j=1$, $2, \cdots, n)$.