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 $$ \left(\sum \limits_{k=1}^n (2k-1)x_k\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{1}{x_k}\right) \le n^2 \left(\sum \limits_{k=1}^n x_k\right) \left( \sum \limits_{k=1}^n \frac{1}{x_k}\right)$$

We have$$\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}{=\sum \limits_{k=1}^n (2k-1)\left\{1+{1\over k}\right\}\\=\sum \limits_{k=1}^n 2k-1+\sum \limits_{k=1}^n {2k-1\over k}\\=n^2+2n-\sum_{k=1}^{n}{1\over k}\\=n^2+n-H_n}$$where $H_n$ is the famous Harmonic Number. Similarly$$\sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}=n^2-2n-3-3H_{n+1}$$and$$n^2 \left(\sum \limits_{k=1}^n \frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n \frac{k}{k+1}\right)=n^2(n+H_n)(n+1-H_{n+1})$$which by substituting, leave us with the following inequality:

$$(n^2+n-H_n)(n^2-2n-3-3H_{n+1})\le n^2(n+H_n)(n+1-H_{n+1})$$

after expanding the terms we obtain$$LHS\le RHS$$where$$LHS=n^4-4n^2+3n^2H_{n+1}+6nH_{n+1}-3n^2-6n-n^2H_{n}+2nH_{n}-3H_{n}H_{n+1}+3H_{n}\\RHS=n^4+n^3-n^3H_{n+1}+n^3H_{n}+n^2H_{n}-n^2H_{n}H_{n+1}$$by rearranging the terms we can rewrite
$$-n^3-7n^2-6n+H_n(-n^3-2n^2+2n+3)+H_{n+1}(n^3+3n^2+6n)+H_nH_{n+1}(n^2-3)\le0$$
or equivalently 

$$3H_n+H_{n+1}(n^2+8n)+H_nH_{n+1}(n^2-3)\le n^3+6n^2+5n+{3n\over n+1}$$

For $n=1$ and $n=2$, this inequality obviously holds and for $n\ge 3$ we divide the two sides of the inequality to $n^2-3$ therefore 

$${3H_n\over n^2-3}+H_{n+1}{n^2+8n\over n^2-3}+H_nH_{n+1}\le n+6+{8\over n}+{21n+24\over n^2-3}$$ 

since for $n\ge 3$ we have ${3H_n\over n^2-3}<1$ we only need to show that

$$H_{n+1}{n^2+8n\over n^2-3}+H_nH_{n+1}\le n+5+{8\over n}+{21n+24\over n^2-3}$$

Also this inequality holds for $3\le n\le 8$ by simple hand calculations and for $n\ge 9$:$${n^2+8n\over n^2-3}<2$$and we only show that

$$2H_{n+1}+H_nH_{n+1}\le n+5+{8\over n}$$

which is obvious since the RHS and LHs are convex and concave functions respectively and the inequality holds for $n=9$. Therefore the proof is complete $\blacksquare$
