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