Interesting four-sum inequality $n^2 \left(\sum \limits_{k=1}^n \frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n \frac{k}{k+1}\right) \ge...$

(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.

Just a partial approach.

For the different sums, we have $$\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}=n^2+2 n-H_n$$ $$\sum \limits_{k=1}^n (2k-1)\frac k{k+1}=3 H_{n+1}+(n-3) (n+1)$$ $$\sum \limits_{k=1}^n \frac{k+1}{k}=H_n+n$$ $$\sum \limits_{k=1}^n \frac k{k+1}=n+1-H_{n+1}$$

So, we need to prove that $$f(n)=n^2 \left(H_n+n\right) \left(n+1-H_{n+1}\right)-\left(n (n+2)-H_n\right) \left(3 H_{n+1}+(n-3) (n+1)\right)\geq 0$$

At least, for large value of $n$, we have $$f(n)=n^3-n^2 \left(\log \left({n}\right)+\gamma -2\right) \left(\log \left({n}\right)+\gamma +3\right)+O(n)$$

For small values of $n$ $$f(n)=\frac{1}{12} \left(120-22 \pi ^2+\pi ^4\right) n^2+O\left(n^3\right)$$