A simple inequality for sinus

Question

Is there a simple proof os this inequality $$\sin x \sqrt{1+\frac{1}{x^2}}\ge 1, \ \ 0\le x\le \pi/2.$$ I have very complicated proof, but seems that maybe there is a simple proof.

Answer 1

Note that $\sin(x)=\frac{\tan(x)}{\sqrt{1+\tan^2(x)}}$ (you can derive this from the definition of $\tan(x)$ and the fundamental identity $1=\cos^2(x)+\sin^2$(x)) for every $x \in [0,\pi/2)$, hence this can be rearranged as: $$ \frac{\tan(x)}{\sqrt{1+\tan^2(x)}} \frac{\sqrt{1+x^2}}{x} \ge 1 \Longleftrightarrow \frac{\tan(x)}{\sqrt{1+\tan^2(x)}} \ge \frac{x}{\sqrt{1+x^2}}$$ Now, let $g(x)=\frac{x}{\sqrt{1+x^2}}$, note that $g$ in increasing in $[0,+\infty)$ and $\tan(x) \ge x $ for every $x \in [0,\pi/2)$. The claim follows (if you want, verify the claim for $x=\pi/2$ where the first identity does not hold true).