First part : The statement in the title is wrong.
Let's suppose that the sequence $(b_k)_{k \geq 1}$ is the following :
$$\dfrac{1}{\ln(2)}, \dfrac{1}{\ln(2)}, \dfrac{1}{\ln(3)}, \dfrac{1}{\ln(3)}, \dfrac{1}{\ln(3)}, \dfrac{1}{\ln(4)}, \dfrac{1}{\ln(4)}, \dfrac{1}{\ln(4)}, \dfrac{1}{\ln(4)}, ...$$
(the $\dfrac{1}{\ln(k)}$ term appearing $k$ times).
Let $N_1=0$ and for any $j \geq 2$, let $N_j = 2+3+...+j$. Then
\begin{align*}\sum_{k=1}^{N_j} \dfrac{b_k}{N_j+1-k} &= \sum_{i=2}^j \sum_{k=N_{i-1}+1}^{N_i} \dfrac{b_k}{N_j+1-k} \\
&= \sum_{i=2}^j \sum_{k=N_{i-1}+1}^{N_i} \dfrac{1}{\ln(i)(N_j+1-k)} \\
&= \sum_{i=2}^j \dfrac{1}{\ln(i)}\sum_{k=N_{i-1}+1}^{N_i} \dfrac{1}{N_j+1-k} \\
&\geq \dfrac{1}{\ln(j)}\sum_{i=2}^j \sum_{k=N_{i-1}+1}^{N_i} \dfrac{1}{N_j+1-k} \\
&=\dfrac{1}{\ln(j)}\sum_{k=1}^{N_j} \dfrac{1}{N_j+1-k} \\
&=\dfrac{1}{\ln(j)}\sum_{k=1}^{N_j} \dfrac{1}{k} \\
&\geq \dfrac{1}{\ln(j)}\sum_{k=1}^{j} \dfrac{1}{k}
\end{align*}
Since $\displaystyle\lim_{j \rightarrow +\infty} \dfrac{1}{\ln(j)}\sum_{k=1}^{j} \dfrac{1}{k} = 1$, then $\displaystyle\sum_{k=1}^{N_j} \dfrac{b_k}{N_j+1-k} $ does not tend to $0$ as $j$ tends to $+\infty$.
Therefore, $\displaystyle\sum_{k=1}^{n} \dfrac{b_k}{n+1-k}$ does not tend to $0$ as $n$ tends to $+\infty$.
Second part : Lets' prove that if $\displaystyle\lim_{n \rightarrow +\infty} a_n=a$, then $\displaystyle\lim\limits_{n \to \infty} \sum\limits_{k=1 }^n \frac{a_k}{(n+1-k)(n+2-k)} = a$.
Let's suppose that $\displaystyle\lim_{n \rightarrow +\infty} a_n=a$, and let $\varepsilon > 0$.
By definition of convergence, there exists $n_0 \in \mathbb{N}$ such that for every $n > n_0$, one has $|a_n-a| \leq \varepsilon$.
Let $M=\max\lbrace |a_0-a|, ..., |a_{n_0}-a| \rbrace$.
Finally, let $N \geq n_0$ such that for every $n \geq N$,
$$\dfrac{|a|}{n+1} + M \left[\frac{1}{n+1-n_0}-\frac{1}{n+1}\right] \leq \varepsilon$$
For every $n > 0$, one has
$$\sum\limits_{k=1 }^n \frac{a}{(n+1-k)(n+2-k)} = a \sum\limits_{k=1 }^n \frac{1}{n+1-k}-\frac{1}{n+2-k} = a \left(1-\dfrac{1}{n+1}\right) $$
so $$a=\dfrac{a}{n+1} +\sum\limits_{k=1 }^n \frac{a}{(n+1-k)(n+2-k)}$$
One deduces that for every $n \geq N$,
\begin{align*} &\left|a-\sum\limits_{k=1 }^n \frac{a_k}{(n+1-k)(n+2-k)}\right| \\ \leq &\left|\dfrac{a}{n+1} + \sum\limits_{k=1 }^n \frac{a-a_k}{(n+1-k)(n+2-k)}\right| \\
\leq &\dfrac{|a|}{n+1} + \sum\limits_{k=1 }^n \frac{|a_k-a|}{(n+1-k)(n+2-k)} \\
\leq &\dfrac{|a|}{n+1} + \sum\limits_{k=1 }^{n_0} \frac{|a_k-a|}{(n+1-k)(n+2-k)}+ \sum\limits_{k=n_0+1 }^{n} \frac{|a_k-a|}{(n+1-k)(n+2-k)} \\
\leq &\dfrac{|a|}{n+1} + \sum\limits_{k=1 }^{n_0} \frac{M}{(n+1-k)(n+2-k)}+ \sum\limits_{k=n_0+1 }^{n} \frac{\varepsilon}{(n+1-k)(n+2-k)} \\
\leq &\dfrac{|a|}{n+1} + M \sum\limits_{k=1 }^{n_0} \left[\frac{1}{n+1-k}-\frac{1}{n+2-k}\right]+ \varepsilon\sum\limits_{k=n_0+1 }^{n} \left[\frac{1}{n+1-k}-\frac{1}{n+2-k}\right] \\
\leq &\dfrac{|a|}{n+1} + M \left[\frac{1}{n+1-n_0}-\frac{1}{n+1}\right]+ \varepsilon \left[1-\frac{1}{n+1-n_0}\right] \\
\leq & \varepsilon + \varepsilon \leq 2 \varepsilon
\end{align*}
which finally proves that $\boxed{\displaystyle\lim\limits_{n \to \infty} \sum\limits_{k=1 }^n \frac{a_k}{(n+1-k)(n+2-k)} = a}$.
Best Answer
This is a classic proof of this weakened Shapiro inequality. Let $x_1$ be the maximum of $a_i,$ $i=1,…,n$. Let $x_2$ be the maximum of the two numbers following $x_1$ in the sequence $\{a_i\}$. Then let $x_3$ be the maximum of the two numbers following $x_2$ and so on. This process will end up back in $x_1$. Then we can say that the LHS of your inequality is greater or equal to
$$\frac{x_1}{2x_2}+ \frac{x_2}{2x_3}+\dots+ \frac{x_m}{2x_1}$$
which is obviously (due to AM-GM) $\geq m/2 \geq n/4$.