If $a_{n+1} \leq \left(1-\frac{2}{n+1}\right)a_n + b\left(\frac{2}{n+1}\right)^2$ then $a_n \leq \frac{4b}{n+1}$ A sequence $(a_n)$ satisfy $a_{n+1} \leq (1-\gamma_n)a_n + \frac{\beta R^2}{2}\gamma_n^2$ where $\gamma_n = \frac{2}{n+1}$, $\beta$ and $R$ are constant.
How to verify that
$$a_n \leq \frac{2\beta R^2}{n+1},$$ note that it is easy to show $a_2\leq\frac{\beta R^2}{2}$

It is easy to show that $a_2 \leq \frac{\beta R^2}{2}$
We use induction to prove the inequality. Suppose that for $i = 2\ldots n$, we have $a_i \leq \frac{2\beta R^2}{i+1}$, then
$$a_{n+1}\leq (1-\gamma_n)a_n + \frac{\beta R^2}{2}\gamma_n^2 = \frac{n-1}{n+1}a_n+\frac{2\beta R^2}{(n+1)^2} \leq \frac{2(n-1)\beta R^2}{(n+1)^2}+\frac{2\beta R^2}{(n+1)^2}=\frac{2n\beta R^2}{(n+1)^2}<\frac{2\beta R^2}{n+2}$$
The last inequality simply based on $\frac{n}{(n+1)^2}\leq \frac{1}{n+2}\Leftrightarrow n^2+2n \leq n^2 + 2n + 1$
So we can conclude the proof.
