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$

Here is a way, which is motivated by the excellent approach of @MichaelRozenberg's. Let us write $\displaystyle a_k = \frac { k}{k+1}, b_k = 2k-1$. We then have \begin{align*} S =& \sum_{i\le n} a_i b_i\sum_{j\le n}\frac{b_j}{a_j} \\ =& \sum_{i,j\le n} \frac{a_i}{a_j}b_ib_j \\ =& \frac 1 2 \sum_{i,j\le n} \left(\frac {a_i}{a_j} + \frac{a_j}{a_i}\right) b_ib_j. \end{align*} We then observe \begin{align*} \frac {a_i}{a_j} + \frac{a_j}{a_i} \le& \begin{cases} 2\quad \text {if}\quad i=j=1\\ \frac{5}{2}\quad\text{if}\quad i=1, j> 1\ \text{ or }\ i>1, j=1\\ \frac{13}{6}\quad \text{if} \quad i > 1, j> 1 \end{cases} \end{align*} thus \begin{align*} S \le &\frac 1 2 \left[2 + 2\cdot\frac {5}{2}(n^2-1) + \frac {13}{6}(n^2-1)^2\right]\\ <& \frac 9 8 n^4 \end{align*} for all $n\ge 1$ except for $n=2$. We can check $n=2$ case manually...

We can use also the Cassel's inequality:

Let $a$, $b$ and $w$ be sequences of $n$ positive numbers such that $1<m\leq\frac{a_k}{b_k}\leq M$ for any $k.$ Prove that: $$\sum_{k=1}^nw_ka_k^2\sum_{k=1}^nw_kb_k^2\leq\frac{(M+m)^2}{4Mm}\left(\sum_{k=1}^nw_ka_kb_k\right)^2.$$

This inequality was here:

G.S. WATSON, Serial Correlation in Regression Analysis, Ph.D. Thesis, Dept. of Experimental Statistics, North Carolina State College, Raleigh; Univ. of North Carolina, Mimograph Ser., No. 49, 1951, appendix 1.

In our case $w_k=2k-1$, $a_k=\sqrt{\frac{k+1}{k}}$, $b_k=\sqrt{\frac{k}{k+1}},$ $M=2$ and $m=1$, which gives: $$\sum_{k=1}^n(2k-1)\frac{k+1}{k}\sum_{k=1}^n(2k-1)\frac{k}{k+1}\leq\frac{(2+1)^2}{4\cdot2\cdot1}\left(\sum_{k=1}^n(2k-1)\right)^2=\frac{9n^4}{8}.$$