Is $\sin\frac{1}{x} \lt \frac{1}{x},\ \forall x\geq 1$?
I tried "copying" the proof of $\ln x \lt x, \forall x\geq 1$ but it didn't quite work.
Here's what I did: Let $f(x)=\sin\frac{1}{x}-\frac{1}{x}$, I'd like to show that $f(x)\lt 0$. f'(x)=\frac{1}{x^2}(1-\cos\frac{1}{x}) \gt 0. So $f(x)$ is increasing, then $f(x)\gt f(1), \forall x \gt 1$ but $f(1)=\sin 1 - 1\lt 0$. And now I'm stuck, not sure if what I did is in the right direction.