<s>[INST] 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.
 [/INST] $\sqrt{t+1}$ is approximately constant on $(0,1)$, hence the following application of Cauchy-Schwarz
$$ I_n=\int_{0}^{1}t^n\sqrt{1+t}\,dt\leq \sqrt{\int_{0}^{1}t^{n}\,dt\int_{0}^{1}t^n(1+t)\,dt}=\color{red}{\frac{1}{n+1}\sqrt{\frac{2n+3}{n+2}}}<\frac{\sqrt{2}}{n+1}$$
is expected to provide a tight upper bound. Indeed,
$$ \frac{\sqrt{2}}{n+1}-I_n = \int_{0}^{1}t^n\left(\sqrt{2}-\sqrt{1+t}\right)\,dt = \int_{0}^{1}(1-t)^n\left(\sqrt{2}-\sqrt{2-t}\right)\,dt $$
leads to
$$\begin{eqnarray*}\frac{\sqrt{2}}{n+1}-I_n = \int_{0}^{1}\frac{t(1-t)^n}{\sqrt{2}+\sqrt{2-t}}\,dt&\leq&\frac{1}{\sqrt{2}+1}\int_{0}^{1}t(1-t)^n\,dt\\&=&\frac{1}{\sqrt{2}+1}\cdot\frac{1}{(n+1)(n+2)}\\&<&\frac{1}{2(n+1)^2}.\end{eqnarray*}$$
Actually we have
$$ I_n = \frac{\sqrt{2}}{n+1}+\frac{C_n}{(n+1)^2} $$
where $C_n\in\left[\frac{1}{2\sqrt{2}},\frac{1}{\sqrt{2}+1}\right]$.
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