Let $m=\cot^2\alpha$ where $\alpha\in(0, \frac{\pi}{2})$ . Then $$\frac{1}{\sqrt{m+1}}<\sin\frac{1}{\sqrt{m}}\Leftrightarrow\sin\alpha < \sin\tan\alpha$$ By # 3892, this relation will hold if $0 <\tan\alpha < \frac{\pi}{2}$. This is indeed the case because $$\cot^2\alpha = m \ge 1 \implies \tan\alpha \le 1<\frac{\pi}{2}$$