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$ 
Prove that for all $n \in \mathbb{N}$ the inequality $$ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le \frac{9}{8}n^4$$ 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 $n=1,2,3,...,15$). I hope that there is nice proof of this inequality. Michael Rozenberg wrote a very nice solution to a similar problem ( Prove the inequality $\sum \limits_{k=1}^n \frac{k+1}{k} \cdot \sum \limits_{k=1}^n \frac{k}{k+1} \le \frac{9}{8}n^2$ ). I think this inequality has a similar proof, but I can’t prove in a similar way. I will write as I proved the inequality. Let $S_n= \sum \limits_{k=1}^n \frac{1}{k} $. Then $$ \sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}=n^2+2n-S_n $$ and $$\sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}=n^2-2n-3+\frac{3}{n+1}+3S_n$$ We need to prove that $$3S_n^2-S_n \left( 2n^2+8n+3-\frac{3}{n+1}\right)+\frac{n^4}{8}+7n^2+3n-3+\frac{3}{n+1} \ge 0$$ To prove this inequality, I found discriminant of the quadratic polynomial and used the fact that $S_n \le n$. It was possible to prove that the inequality holds for all $n \ge 16$.
 A: 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...
A: 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:
$$\sum_{k=1}^n(2k-1)\frac{k+1}{k}\sum_{k=1}^n(2k-1)\frac{k}{k+1}\leq\frac{(2+1)^2}{4\cdot2\cdot1}\left(\sum_{k=1}^n(2k-1)\right)^2=\frac{9n^4}{8}.$$
