Interesting four-sum inequality $n^2 \left(\sum \limits_{k=1}^n \frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n \frac{k}{k+1}\right) \ge...$ 
Prove that for all $n \in \mathbb{N}$ the inequality $$ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le n^2 \left(\sum \limits_{k=1}^n \frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n \frac{k}{k+1}\right)$$ holds.

My work.In fact, I want to solve another problem ( Prove $ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le \frac{9}{8}n^4$ ) This another problem has already been solved, but I want to solve it by the method that the author of the problem intended. The fact is that in the original problem there was another inequality ( Prove the inequality $\sum \limits_{k=1}^n \frac{k+1}{k} \cdot \sum \limits_{k=1}^n \frac{k}{k+1} \le \frac{9}{8}n^2$ ). It seems to me that the inequality $\sum \limits_{k=1}^n \frac{k+1}{k} \cdot \sum \limits_{k=1}^n \frac{k}{k+1} \le \frac{9}{8}n^2$ is proved as the author of the problem wanted. I think that the author of the problem wanted us to prove inequality $ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le \frac{9}{8}n^4$ on the basis of inequality $\sum \limits_{k=1}^n \frac{k+1}{k} \cdot \sum \limits_{k=1}^n \frac{k}{k+1} \le \frac{9}{8}n^2$. To this end, it suffices to prove the inequality $$ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le n^2 \left(\sum \limits_{k=1}^n \frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n \frac{k}{k+1}\right)$$ I checked this inequality by numerical methods on a computer. This inequality holds for all $n$. Interestingly, if $n \to \infty$ then this inequality (in the limit) is an equality. 
Perhaps this will help to solve the problem: let $x_k=\frac{k+1}{k}$. Then $1<x_k \le 2$ and inequality takes the form $$ \left(\sum \limits_{k=1}^n (2k-1)x_k\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{1}{x_k}\right) \le n^2 \left(\sum \limits_{k=1}^n x_k\right) \left( \sum \limits_{k=1}^n \frac{1}{x_k}\right)$$

Here's a probabilistic interpretation and solution to this problem, although the method itself may be negligibly faster than the existing answers. Let $X$ be a discrete random uniform random variable on $\{1, \cdots, n\}$. Then the problem can be rewritten as
\begin{align*}
\mathbb{E}\left[(2X-1)\frac{X+1}{X}\right]\mathbb{E}\left[(2X-1)\frac{X}{X+1}\right] = \mathbb{E}[2X-1]\mathbb{E}\left[\frac{X+1}{X}\right]\mathbb{E}[2X-1]\mathbb{E}\left[\frac{X}{X+1}\right]
\end{align*}
Rewritting the LHS in terms of $\mathbb{E}[f(X)g(X)] = \text{Cov}(f(X), g(X)) + \mathbb{E}[f(X)]\mathbb{E}[g(X)]$ and dividing by the RHS,
\begin{align*}
\left(1 + \frac{\text{Cov}\left(2X-1, \frac{X+1}{X}\right)}{n\mathbb{E}\left[\frac{X+1}{X}\right]}\right)\left(1 + \frac{\text{Cov}\left(2X-1, \frac{X}{X+1}\right)}{n\mathbb{E}\left[\frac{X}{X+1}\right]}\right) \le 1
\end{align*}
Note that
\begin{align*}
\text{Cov}\left(2X-1, \frac{X+1}{X}\right) = 2\text{Cov}\left(X, \frac{X+1}{X}\right) = n+3 - (n+1)\mathbb{E}\left[\frac{X+1}{X}\right] \\
\text{Cov}\left(2X-1, \frac{X}{X+1}\right) = 2\text{Cov}\left(X+1, \frac{X}{X+1}\right) = n+1 - (n+3)\mathbb{E}\left[\frac{X}{X+1}\right]
\end{align*}
So we have to prove
\begin{align*}
\left(-\frac{1}{n} + \left(1 + \frac{3}{n}\right)\frac{1}{\mathbb{E}[\frac{X+1}{X}]}\right)\left(-\frac{3}{n} + \left(1 + \frac{1}{n}\right)\frac{1}{\mathbb{E}[\frac{X}{X+1}]}\right) \le 1
\end{align*}
Let $a_n = \mathbb{E}[\frac{X+1}{X}]$ and $b_n = \mathbb{E}[\frac{X}{X+1}]$. Reorganizing, the statement we need to prove is
\begin{align*}
f(n) = (a_n b_n - 1)n^2 + (3b_n + a_n - 4)n + (a_n + 9b_n - 3a_nb_n - 3) \ge 0
\end{align*}
It is straightforward to prove the bounds
\begin{align*}
1 &< a_n = 1 + \frac{H_n}{n} \\
1 + \frac{1}{n+1} - \frac{H_n}{n} &= b_n < 1 \\
a_n b_n &= 1 - \text{Cov}\left(\frac{X+1}{X}, \frac{X}{X+1}\right) \\
&= 1 + \text{Cov}\left(\frac{1}{X}, \frac{1}{X+1}\right) \\
&= 1 + \mathbb{E}\left[\frac{1}{X(X+1)}\right] - \mathbb{E}\left[\frac{1}{X}\right]\mathbb{E}\left[\frac{1}{X+1}\right] \\
&= 1 + \frac{1}{n}\left(1 - \frac{1}{n+1}\right) - \mathbb{E}\left[\frac{1}{X}\right]\mathbb{E}\left[\frac{1}{X+1}\right] \\
&\begin{cases} < 1 + \frac{1}{n}  \\
> 1 + \frac{1}{n} - \frac{1}{n(n+1)} - \frac{H^2_n}{n^2} 
\end{cases}
\end{align*}
Therefore,
\begin{align*}
f(n) &> \left(\frac{1}{n} - \frac{1}{n(n+1)} - \frac{H^2_n}{n^2}\right)n^2 + \left(\frac{3}{n+1} - 2\frac{H_n}{n}\right)n + \left(4 + \frac{9}{n+1} - \frac{8H_n}{n} - \frac{3}{2n}\right) \\
&= n-\frac{n}{n+1} - H_n^2 + 3\frac{n}{n+1} - 2H_n + \left(4 + \frac{9}{n+1} - \frac{8H_n}{n} - \frac{3}{2n}\right) \\
&= n + \underbrace{2\frac{n}{n+1}}_{\ge 1.9 \text{ for } n \ge 20} - H_n^2 - 2H_n + \underbrace{\left(4 + \frac{9}{n+1} - \frac{8H_n}{n} - \frac{3}{2n}\right)}_{\ge 2.9 \text{ for } n \ge 20} \\
&> n + 4.8-2H_n - H_n^2
\end{align*}
and the last expression is $> 0$ and increasing for $n \ge 20$. We can manually check the inequality is satisfied for $1 \le n \le 19$. With some better bounding, we can decrease the number of cases we need to check.
