Showing that the inequality $\frac{\sqrt{2}}{n+1} - \frac{1}{2n^2} \leq u_n \leq \frac{\sqrt{2}}{n+1}$ stands. 
If $\displaystyle u_{n}=\int_{0}^{1} t^n \sqrt{t+1} dt$, where $n\geq 1$, prove that 
  $$\frac{\sqrt{2}}{n+1}  - \frac{1}{2n^2} \leq u_n \leq  \frac{\sqrt{2}}{n+1}.$$

I have already proven the second inequality, but couldn't the first one. I managed to show that $\dfrac{1}{n+1}  \leq u_n$, and I wrote a recursive formula by using the $\displaystyle \int_{0}^{1} \dfrac{t^n}{\sqrt{t+1}} dt$ integral, it got worse. Seems to be some easier way.

Since $t\to\sqrt{t+1}$ is concave, it follows that, for $t\in [0,1]$,
$$1+(\sqrt{2}-1)t\leq \sqrt{t+1}\leq \sqrt{2}.$$
Hence 
$$\frac{1}{n+1}+\frac{\sqrt{2}-1}{n+2}\leq \int_{0}^{1} t^n \sqrt{t+1} dt\leq \frac{\sqrt{2}}{n+1}.$$
It remains to show that
$$
\frac{\sqrt{2}}{n+1}  - \frac{1}{2n^2}\leq
\frac{1}{n+1}+\frac{\sqrt{2}-1}{n+2}$$
that is
$$\frac{\sqrt{2}-1}{(n+1)(n+2)}\leq \frac{1/2}{n^2}$$
which holds for all $n\geq 1$.
