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 $f(n)\ge f(n+1)$?
 A: The first is true because $n\leq i(n+1-i)$ for all $1\leq i\leq n$.
Rewrite $f(n)=\frac2{n+1}\left[1+\frac12+\frac13+\cdots+\frac1n\right]$.
Then, in the following difference, $1/k$ is multiplied by $\frac2{n+1}-\frac2{n+2}$, leaving the final $\frac1{n+1}$ multiplied by $\frac2{n+2}$.
$$f(n)-f(n+1)=\frac2{(n+1)(n+2)}\left[1+\frac12+...+\frac1n\right]-\frac2{(n+1)(n+2)}>0$$
A: Note that
$i \leq n$
$i(i-1) \leq n(i-1)$
$i^2-i \leq ni-i^2+i$
$n \leq ni-i^2+i$
$\frac{1}{n} \geq \frac{1}{i(n-i+1)} $
Since the series has $n$ terms and each term is less than $1/n$, the sum of series is less than $1$.
A: 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 $1 \leq i \leq n$
A: $\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
$$1
\ge \sum_{i=1}^n \frac1{i(n+1-i)}
\ge \frac{4n}{(n+1)^2}
.$$
