Prove $\frac{n}{\sum_{k=1}^n{\frac{1}{\frac{1}{k}+a_k}}}-\frac{n}{\sum_{k=1}^n{\frac{1}{a_k}}}\geqslant \frac{2}{n+1}$

I gave a solution years ago.

Denote $A = \sum_{k=1}^n \frac{1}{\frac{1}{k} + a_k}$, $B = \sum_{k=1}^n \frac{1}{a_k}$.

The function $f(x) = \frac{1}{1 + \frac{1}{x}}, \ x > 0$ is concave. By Jensen's inequality, we have \begin{align} A &= f(\tfrac{1}{a_1}) + 2f(\tfrac{1}{2a_2}) + 3(\tfrac{1}{3a_3}) + \cdots + n f(\tfrac{1}{na_n})\\ &\le \frac{n(n+1)}{2} f\Big( \frac{B}{\frac{n(n+1)}{2}}\Big)\\ &= \frac{n(n+1)}{2}\frac{B}{B + \frac{n(n+1)}{2}}. \end{align} Then we have $$\frac{n}{A} - \frac{n}{B} \ge \frac{n}{\frac{n(n+1)}{2}\frac{B}{B + \frac{n(n+1)}{2}}} - \frac{n}{B} = \frac{2}{n+1}\frac{B + \frac{n(n+1)}{2}}{B} - \frac{n}{B} = \frac{2}{n+1}.$$ We are done.