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 know the question has already been answered, and the answer is quite nice, but I was thinking along a different line:
$$\sum\limits_{k=1}^{n} \frac{\frac{1}{k}}{a_k\left(\frac{1}{k}+a_k\right)} \geq \frac{2}{n(n+1)}\left(\sum\limits_{k=1}^{n}\frac{1}{a_k}\right)\left(\sum\limits_{k=1}^{n}\frac{1}{\frac{1}{k}+a_k}\right)$$
Since $\displaystyle \sum\limits_{k=1}^{n}k = \dfrac{n(n+1)}{2}$, we have
$$\left(\sum\limits_{k=1}^{n} k \right)\left(\sum\limits_{k=1}^{n} \frac{\frac{1}{k}}{a_k\left(\frac{1}{k}+a_k\right)}\right) \geq \left(\sum\limits_{k=1}^{n}\frac{1}{a_k}\right)\left(\sum\limits_{k=1}^{n}\frac{1}{\frac{1}{k}+a_k}\right)$$
$$\left(1+2+3+ \cdot \cdot \cdot +n\right)\cdot \left(\frac{1}{a_1(1+a_1)}+\frac{\frac{1}{2}}{a_2(\frac{1}{2}+a_2)}+\cdot\cdot\cdot + \frac{\frac{1}{n}}{a_n(\frac{1}{n}+ a_n)} \right) \geq \\ \left(\frac{1}{a_1}+\frac{1}{a_2}+\cdot\cdot\cdot+\frac{1}{a_n}\right)\cdot\left(\frac{1}{1+a_1}+\frac{1}{\frac{1}{2}+a_2}+\cdot\cdot\cdot+\frac{1}{\frac{1}{n}+a_n}\right)$$
These multiplications can be expanded into sums which can be organized into rows and columns, like an $n \times n$ matrix, say '$B\geq C$'
$$
\frac{1}{a_1(1+a_1)}+\frac{\frac{1}{2}}{a_2(\frac{1}{2}+a_2)}+\cdot\cdot\cdot + \frac{\frac{1}{n}}{a_n(\frac{1}{n}+ a_n)}\\
\frac{2}{a_1(1+a_1)}+\frac{1}{a_2(\frac{1}{2}+a_2)}+\cdot\cdot\cdot + \frac{\frac{1}{n}}{a_n(\frac{2}{n}+ a_n)}
\\
\vdots\\
\frac{n}{a_1(1+a_1)}+\frac{\frac{n}{2}}{a_2(\frac{1}{2}+a_2)}+\cdot\cdot\cdot + \frac{1}{a_n(\frac{1}{n}+ a_n)}\\
\geq \\
\frac{1}{a_1(1+a_1)}+\frac{1}{a_1(\frac{1}{2}+a_2)}+\cdot\cdot\cdot + \frac{1}{a_1(\frac{1}{n}+ a_n)}\\
\frac{1}{a_2(1+a_1)}+\frac{1}{a_2(\frac{1}{2}+a_2)}+\cdot\cdot\cdot + \frac{1}{a_2(\frac{1}{n}+ a_n)}\\
\vdots\\
\frac{1}{a_n(1+a_1)}+\frac{1}{a_n(\frac{1}{2}+a_2)}+\cdot\cdot\cdot + \frac{1}{a_n(\frac{1}{n}+ a_n)}
$$
Now note that the diagonals are equal, so we don't need to worry about them.
However, take for example the opposite entries of the diagonals,
First the entries in the first 'matrix' $B$ are of the form $b_{ij}=\dfrac{\frac{i}{j}}{a_j\left(\frac{1}{j}+a_j\right)}$ while the entries in the second 'matrix' $C$ are of the form $c_{ij}=\dfrac{1}{a_i\left(\frac{1}{j}+a_j\right)}$
then $c_{ij}+c_{ji} \leq b_{ij} + b_{ji}$:
$$\frac{1}{a_i\left(\frac{1}{j}+a_j\right)}+\frac{1}{a_j\left(\frac{1}{i}+a_i\right)} \leq \frac{\frac{i}{j}}{a_j\left(\frac{1}{j}+a_j\right)}+\frac{\frac{j}{i}}{a_i\left(\frac{1}{i}+a_i\right)}\\
a_j\left(\frac{1}{i}+a_i\right)+a_i\left(\frac{1}{j}+a_j\right) \leq \frac{i}{j}a_i\left(\frac{1}{i}+a_i\right)+\frac{j}{i}a_j\left(\frac{1}{j}+a_j\right)\\
\frac{1}{i}a_j+a_i a_j+\frac{1}{j}a_i + a_i a_j \leq \frac{1}{j}a_i+\frac{i}{j}a^{2}_i+\frac{1}{i}a_j+\frac{j}{i}a^{2}_j\\
2ija_i a_j \leq i^2 a^{2}_i+ j^2 a^{2}_j$$
and
$$0 \leq i^2 a^{2}_i - 2ija_i a_j + j^2 a^{2}_j = (ia_i-ja_j)^2$$
the inequalities hold because $a_k>0$ for $k=1,2,...,n$
Therefore the sum of the upper and lower triangular parts of $B$ is $\geq$ to the sum of the of the upper and lower triangular parts of $C$.
Sorry for the extra long answer.
