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 $$ \left(\sum \limits_{k=1}^n (2k-1)x_k\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{1}{x_k}\right) \le n^2 \left(\sum \limits_{k=1}^n x_k\right) \left( \sum \limits_{k=1}^n \frac{1}{x_k}\right)$$
Best Answer
(update 2019/08/09)
Let $H_n = \sum_{k=1}^n \frac{1}{k}$. Since $\frac{1}{k} \le -\ln (1 - \frac{1}{k}) $ for $k\ge 2$, we have $H_n - 1 \le -\sum_{k=2}^n \ln (1 - \frac{1}{k}) = \ln n$ or $H_n \le \ln n + 1$.
Rewrite the inequality as $$(n^2+2n-H_n)\left(n^2 - 2n - 3 + \frac{3}{n+1} + 3H_n\right) \le n^2(n+ H_n)\left(n + 1 - \frac{1}{n+1} - H_n\right)$$ or $$-(n^2-3)H_n^2 - \frac{n(n^2+10n+11)}{n+1}H_n + \frac{n^2(n+2)(n+5)}{n+1}\ge 0. \qquad (1)$$
When $1\le n\le 12$, the inequality in (1) is verified directly.
When $n\ge 13$, rewrite (1) as $$-H_n^2 - \frac{n(n^2+10n+11)}{(n+1)(n^2-3)}H_n + \frac{n^2(n+2)(n+5)}{(n+1)(n^2-3)} \ge 0.$$ Since $\frac{n(n^2+10n+11)}{(n+1)(n^2-3)} \le \frac{7}{4}$ and $\frac{n^2(n+2)(n+5)}{(n+1)(n^2-3)} \ge n+6$, it suffices to prove that $-H_n^2 - \frac{7}{4}H_n + n+6 \ge 0$ or $$-\frac{7}{8} + \frac{1}{8}\sqrt{433+64n} \ge H_n.$$ Since $H_n \le \ln n + 1$, it suffices to prove that $-\frac{7}{8} + \frac{1}{8}\sqrt{433+64n} \ge \ln n + 1$. Let $f(x) = -\frac{7}{8} + \frac{1}{8}\sqrt{433+64x} - \ln x - 1$. We have $f(13) > 0$ and $f'(x) = \frac{4}{\sqrt{433+64x}} - \frac{1}{x} > 0$ for $x\ge 13$. Thus, $f(n) \ge 0$ for $n\ge 13$. This completes the proof.