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 $f(n)\ge f(n+1)$?

For second part
$f(n)-f(n+1)= \sum\limits_{i=1}^n [\frac{1}{i(n+1-i)}-\frac{1}{i(n+2-i)}] -\frac{1}{n+1}$
$= \sum\limits_{i=1}^n [\frac{1}{i(n+1-i)}-\frac{1}{i(n+2-i)} -\frac{1}{n(n+1)}]$
$= \sum\limits_{i=1}^n [\frac{1}{i(n+1-i)}-\frac{1}{i(n+2-i)} -(\frac{1}{n}-\frac{1}{n+1})]$
$= \sum\limits_{i=1}^n [(\frac{1}{i(n+1-i)}-\frac{1}{n})-(\frac{1}{i(n+2-i)} -\frac{1}{(n+1)})]$
$=\sum\limits_{i=1}^n [\frac{n-in-i+i^2}{i.n(n+1-i)}-\frac{n+1-in-2i+i^2}{i(n+1)(n+2-i)}]$
$\leq\sum\limits_{i=1}^n [ \frac{n-in-i+i^2}{i(n+1)(n+2-i)}-\frac{n+1-in-2i+i^2}{i(n+1)(n+2-i)}]$
$=\sum\limits_{i=1}^n \frac{i-1}{i(n+1)(n+2-i)}  \geq0$  ,   as $(i-1) \geq 0$ for $1 \leq i \leq n$
