21. Induction on $n$. When $n=2$, the inequality
$$\frac{a_{1} b_{1}}{a_{1}+b_{1}}+\frac{a_{2} b_{2}}{a_{2}+b_{2}} \leqslant \frac{\left(a_{1}+a_{2}\right)\left(b_{1}+b_{2}\right)}{a_{1}+a_{2}+b_{1}+b_{2}}$$

is equivalent to the inequality
$$\begin{array}{l} {\left[a_{1} b_{1}\left(a_{2}+b_{2}\right)+a_{2} b_{2}\left(a_{1}+b_{1}\right)\right]\left(a_{1}+a_{2}+b_{1}+b_{2}\right)} \\ \leqslant\left(a_{1}+a_{2}\right)\left(b_{1}+b_{2}\right)\left(a_{1}+b_{1}\right)\left(a_{2}+b_{2}\right) . \end{array}$$

After simplification and rearrangement, this inequality becomes
$$\left(a_{1} b_{2}-a_{2} b_{1}\right)^{2} \geqslant 0 .$$

Inductive step:
$$\begin{aligned} \sum_{k=1}^{n+1} \frac{a_{k} b_{k}}{a_{k}+b_{k}} & \leqslant \frac{A^{\prime} B^{\prime}}{A^{\prime}+B^{\prime}}+\frac{a_{n+1} b_{n+1}}{a_{n+1}+b_{n+1}} \\ & \leqslant \frac{A B}{A+B}, \end{aligned}$$

where $A^{\prime}=\sum_{k=1}^{n} a_{k}, B^{\prime}=\sum_{k=1}^{n} b_{k}$.