Prove that $\sum_{i=1}^{i=n} \frac{1}{i(n+1-i)} \le1$ $$f(n)=\sum_{i=1}^{i=n} \dfrac{1}{i(n+1-i)} \le 1$$
For example, we have $f(3)=\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot2}+\dfrac{1}{3\cdot1}=\dfrac{11}{12}\lt 1$
If true, it can be used to prove:

Proving $x\ln^2x−(x−1)^2<0$ for all $x∈(0,1)$

Also, can you prove $f(n)\ge f(n+1)$?

$\begin{array}\\
i(n+1-i)
&=i(n+1)-i^2\\
&=(n+1)^2/4-(n+1)^2/4+i(n+1)-i^2\\
&=(n+1)^2/4-((n+1)/2-i)^2\\
\end{array}
$
Therefore
$$i(n+1-i)
=(n+1)^2/4-((n+1)/2-i)^2
\le (n+1)^2/4
$$
and,
since $1 \le i \le n$,
so that
$(n-1)/2 \ge (n+1)/2-i
\ge -(n-1)/2$,
$$i(n+1-i)
=(n+1)^2/4-((n+1)/2-i)^2
\ge (n+1)^2/4-((n-1)/2)^2
=n
$$
so
$$\frac1{n}
\ge \frac1{i(n+1-i)}
\ge \frac{4}{(n+1)^2}
.$$
Therefore
$$1
\ge \sum_{i=1}^n \frac1{i(n+1-i)}
\ge \frac{4n}{(n+1)^2}
.$$
