How to prove $\sin(\frac{1}x) < \frac{1}x$ for all $x$ greater than $1$. How to prove $\sin(\frac{1}x) < \frac{1}x$ for all $x$ greater than $1$?
I was thinking using slopes but I get a contradiction (i.e. $\cos(\frac{1}x) > 1$) when I do some of the algebra.

Let $x>1$ and consider $$\sin(y) = \sum_{n=0}^\infty \frac{(-1)^ny^{2n+1}}{(2n+1)!}$$  Substitute $y = \frac{1}{x}$: $$\sin\left(\frac{1}{x}\right) = \sum_{n=0}^\infty \frac{(-1)^n}{x^{2n+1}(2n+1)!} = \frac{1}{x}-\frac{1}{3!x^3}+\frac{1}{5!x^5}-\frac{1}{7!x^7}+\frac{1}{9!x^9}- \dots \\ = \frac{1}{x}-\left(\frac{1}{3!x^3}-\frac{1}{5!x^5}\right)-\left(\frac{1}{7!x^7}-\frac{1}{9!x^9}\right)- \dots$$ now observe that each quantity in parentheses is positive, so we can think of $\sin\left(\frac{1}{x}\right)$ as $\frac{1}{x}$ minus an infinite amount of positive quantities. It follows that $$\frac{1}{x}> \frac{1}{x}-\left(\frac{1}{3!x^3}-\frac{1}{5!x^5}\right)-\left(\frac{1}{7!x^7}-\frac{1}{9!x^9}\right)- \dots = \sin\left(\frac{1}{x}\right)$$
