I want some help, please, to solve this problem.
Let
$$\begin{cases}x_0=a, a>1\\x_{n+1}=\frac{x_n}{1+\big(n+1\big)x_n^2}\end{cases}$$
I want to prove that $\forall n\in\mathbb{N}^*,\ 0<x_n\leq\frac{1}{n+1}$
I proved that $x_n$ is bounded from below by $0$, but I can't figure out how to prove that $x_n\leq\frac{1}{n+1}$ .
I used the Mathematical Induction method, so for $n=1$ we have : $x_1=\frac{1}{2}$
Then the statement, in this case, is satisfied. Next, we assume another real non-null number $k/\ x_k⩽\frac{1}{n+1}$, and we prove that $x_{k+1}⩽\frac{1}{n+2}$, but I'm stuck.
So please help me if you can. Thanks!
For these types of expressions, I often find it useful to take the reciprocal.
$x_{n+1}=\frac{x_n}{1+\big(n+1\big)x_n^2} $ becomes $\dfrac1{x_{n+1}} =\dfrac{1+\big(n+1\big)x_n^2}{x_n} =\dfrac1{x_n}+(n+1)x_n $.
Letting $y_n = \dfrac1{x_n}$, this becomes $y_{n+1} =y_n+\dfrac{n+1}{y_n} =\dfrac{y_n^2+n+1}{y_n} $, so we need $y_n^2+n+1 \gt (n+2)y_n $ or $y_n^2-(n+2)y_n+n+1 \gt 0 $ or $y_n^2-2(n/2+1)y_n+(n/2+1)^2-(n/2+1)^2+n+1 \gt 0 $ or $(y_n-(n/2+1))^2-n^2/4-n-1+n+1 \gt 0 $ or $(y_n-(n/2+1))^2 \gt n^2/4 $.
Since $y_n > n+1$, $y_n-(n/2+1) \gt n+1-(n/2+1) =n/2 $, so $(y_n-(n/2+1))^2 \gt n^2/4 $ which is what we want.
Therefore $x_n \lt \dfrac1{n+1}$.