Prove the inequality $\sum \limits_{k=1}^n \frac{k+1}{k} \cdot \sum \limits_{k=1}^n \frac{k}{k+1} \le \frac{9}{8}n^2$ 
Prove that for all $n \in \mathbb{N}$ the inequality $$\sum \limits_{k=1}^n \frac{k+1}{k} \cdot \sum \limits_{k=1}^n \frac{k}{k+1} \le \frac{9}{8}n^2$$ holds.

My work. I proved this inequality, but my proof is ugly (it is necessary to check by brute force whether the inequality holds for $n=1,2,3,4,5$). 
 A: The first inequality is equivalent to
$$(n+H_n)(n+1-H_{n+1})\le\frac98n^2$$
where $H_n$ denotes the $n$th Harmonic number. It's well known that $H_n=\gamma+\ln{(n)}+o(1)$ as $n\to\infty$ so it's clear that the LHS is asymptotic to $n^2+n+o(n)$. In particular, we have that the LHS divided by $n^2$ is decreasing. So we just need to find the ratio of this function and $n^2$ for small $n$ and as the ratio begins to decrease, it will not increase again due to the function's asymptotic behaviour. Using this gives us the much stricter inequality
$$(n+H_n)(n+1-H_{n+1})\le\frac{104273}{100800}n^2$$
for $n\in\mathbb{N}$. Equality only occurs when $n=6$ and the ratio is at a maximum.
The second inequality is equivalent to
$$(n^2+2n-H_n)(n^2-2n-3+3H_{n+1})\le\frac98n^4$$
A similar analysis to before gives us the asymptotic $n^4+n^2(2\ln{(n)}+2\gamma-7)+o(n^2)$ for the LHS. Dividing this by $n^4$ gives a decreasing ratio so we again only need to analyse small values of $n$. This gives us the stricter inequality
$$(n^2+2n-H_n)(n^2-2n-3+3H_{n+1})\le\frac{1975}{1944}n^4$$
for $n\in\mathbb{N}$. Equality only occurs when $n=3$ in this case.
A: First problem.
Let $x_k=\frac{k+1}{k}.$
Thus, $x_k\in[1,2]$ for all $k$.
Now, let $$f(x_1,x_2,...x_n)=\sum_{k=1}^nx_k\sum_{k=1}^n\frac{1}{x_k}.$$
Easy to see that $f$ is a convex function of $x_k$ for all $k$.
Thus, by AM-GM $$\max_{x_k\in[1,2]}f=\max_{x_k\in\{1,2\}}f=\max_{i+j=n}(i+2j)\left(i+\frac{j}{2}\right)=$$
$$=\max_{i+j=n}\left((i+j)^2+\frac{1}{2}ij\right)\leq n^2+\frac{1}{2}\left(\frac{n}{2}\right)^2=\frac{9}{8}n^2.$$
