<s>[INST] How prove this inequality $a_{n}>\frac{2n}{n+2}$ Let sequence $\{a_{n}\}$ such $a_{1}=1$,and such
$$a_{n+1}=a_{n}+\dfrac{a^2_{n}}{(n+1)^2}$$\show that
$$a_{n}>\dfrac{2n}{n+2}\tag{1}$$
I try use induction,then  It's a pity that failed,  becase 
$$a_{n+1}>\dfrac{2n}{n+2}+\dfrac{4n^2}{(n+1)^2(n+2)^2}$$
it need to show 
$$\dfrac{2n}{n+2}+\dfrac{4n^2}{(n+1)^2(n+2)^2}>\dfrac{2(n+1)}{n+3}$$
since
$$\dfrac{2n}{n+2}+\dfrac{4n^2}{(n+1)^2(n+2)^2}-\dfrac{2(n+1)}{n+3}=-\dfrac{4(n^2+5n+2)}{(n+1)^2(n+2)^2(n+3)}$$
so how prove this (1)? Thanks,maybe $(1)$ is easy to prove it.
 [/INST] Note that for any $n \in \mathbb{N}_+$,$$
\frac{2n + 2}{n + 3} > \frac{2n}{n + 2} \Longleftrightarrow (n + 1)(n + 2) > n(n + 3),
$$
thus it suffices to prove that$$
a_n \geqslant \frac{2n + 2}{n + 3}. \quad \forall n \in \mathbb{N}_+ \tag{1}
$$
For $n = 1$, $\displaystyle a_1 = 1 = \frac{2 + 2}{1 + 3}$, so (1) holds. Now suppose (1) holds for $n$, then\begin{align*}
a_{n+ 1} &= a_n + \frac{a_n^2}{(n + 1)^2} \geqslant \frac{2n + 2}{n + 3} + \frac{1}{(n + 1)^2} \left(\frac{2n + 2}{n + 3}\right)^2\\
&= \frac{2n + 2}{n + 3} + \frac{4}{(n + 3)^2} = \frac{2n^2 + 8n + 10}{(n + 3)^2}.
\end{align*}
Because\begin{align*}
\frac{2n^2 + 8n + 10}{(n + 3)^2} \geqslant \frac{2n + 4}{n + 4} &\Longleftrightarrow (n + 4)(n^2 + 4n + 5) \geqslant (n + 2)(n + 3)^2\\
&\Longleftrightarrow n^3 + 8n^2 + 21n + 20 \geqslant n^3 + 8n^2 + 21n + 18,
\end{align*}
then$$
a_{n+ 1} \geqslant \frac{2n^2 + 8n + 10}{(n + 3)^2} \geqslant \frac{2n + 4}{n + 4} = \frac{2(n + 1) + 2}{(n + 1) + 3}.
$$
By induction, (1) holds.
</s>