Problem - 3894
Prove $$\frac{1}{\sqrt{2019}} < \underbrace{\sin\sin\sin\cdots\sin}_{2017}\frac{\sqrt{2}}{2} < \frac{2}{\sqrt{2019}}$$
By # 3891:
\begin{align*}
& \underbrace{\sin\sin\sin\cdots\sin}_{2017}\frac{\sqrt{2}}{2}\\
=\quad & \underbrace{\sin\sin\sin\cdots\sin}_{2017}\frac{1}{\sqrt{2}}\\
>\quad & \underbrace{\sin\sin\sin\cdots\sin}_{2016}\frac{1}{\sqrt{3}}\\
>\quad & \underbrace{\sin\sin\sin\cdots\sin}_{2015}\frac{1}{\sqrt{4}}\\
>\quad & \cdots\\
>\quad & \sin\frac{1}{\sqrt{2018}}\\
>\quad & \frac{1}{\sqrt{2019}}
\end{align*}
Meanwhile, by By %%HREF%%3893%%:
\begin{align*}
& \underbrace{\sin\sin\sin\cdots\sin}_{2017}\frac{\sqrt{2}}{2}\\
=\quad & \underbrace{\sin\sin\sin\cdots\sin}_{2017}\frac{2}{\sqrt{8}}\\
<\quad & \underbrace{\sin\sin\sin\cdots\sin}_{2016}\frac{2}{\sqrt{9}}\\
<\quad & \underbrace{\sin\sin\sin\cdots\sin}_{2015}\frac{2}{\sqrt{10}}\\
<\quad & \cdots\\
<\quad & \sin\frac{2}{\sqrt{2024}}\\
<\quad & \frac{2}{\sqrt{2025}}\\
<\quad & \frac{2}{\sqrt{2019}}
\end{align*}