Prove $\frac{n}{\sum_{k=1}^n{\frac{1}{\frac{1}{k}+a_k}}}-\frac{n}{\sum_{k=1}^n{\frac{1}{a_k}}}\geqslant \frac{2}{n+1}$ Let $a_k>0,k=1,2,\cdots, n$. Prove that
$$
\frac{n}{\sum_\limits{k=1}^n{\frac{1}{\frac{1}{k}+a_k}}}-\frac{n}{\sum_\limits{k=1}^n{\frac{1}{a_k}}}\geqslant \frac{2}{n+1}
$$
My attempt: multiply both sides by the denominators, and it is equivalent to prove:
$$
n\sum_{k=1}^n{\frac{1}{a_k}}-n\sum_{k=1}^n{\frac{1}{\frac{1}{k}+a_k}}\geqslant \frac{2}{n+1}\left( \sum_{k=1}^n{\frac{1}{a_k}} \right) \left( \sum_{k=1}^n{\frac{1}{\frac{1}{k}+a_k}} \right) \\
\sum_{k=1}^n{\frac{1}{a_k}}-\sum_{k=1}^n{\frac{1}{\frac{1}{k}+a_k}}\geqslant \frac{2}{n\left( n+1 \right)}\left( \sum_{k=1}^n{\frac{1}{a_k}} \right) \left( \sum_{k=1}^n{\frac{1}{\frac{1}{k}+a_k}} \right)\\ \sum_{k=1}^n{\left( \frac{1}{a_k}-\frac{1}{\frac{1}{k}+a_k} \right)}\geqslant \frac{2}{n\left( n+1 \right)}\sum_{k=1}^n{\frac{1}{a_k}}\sum_{p=1}^n{\frac{1}{\frac{1}{p}+a_p}}
\\ \sum_{k=1}^n{\left( \frac{\frac{1}{k}}{a_k\left( \frac{1}{k}+a_k \right)} \right)}\geqslant \frac{2}{n\left( n+1 \right)}\sum_{k=1}^n{\sum_{p=1}^n{\frac{1}{a_k\left( \frac{1}{p}+a_p \right)}}}
$$
But how to proceed further?

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.
