# Prove the inequality $n \sin \frac{\alpha}{n} \le m\tan \frac{\alpha}{m}$ for $m,n\in\mathbb{N}$ and $\alpha \in (0,\pi/2)$.

I have the following problem:

Let $m,n$ be two positive integers and $\alpha$ an angle between $0$ and $\frac{\pi}{2}$. Prove that:

$n \sin\left(\frac{\alpha}{n}\right) \le m \tan\left(\frac{\alpha}{m}\right)$

I tried solving this problem the following way but I am not sure if it's correct:

$0\le\alpha\le\frac{\pi}{2}$

$0\le\sin (\alpha)\le1$

Since the argument is $\alpha$/n and n is a positive integer, then the sine would have a smaller value but it is still true to say that:

$0\le \sin \left(\frac{\alpha}{n}\right) \le 1$

$0\le n \sin \left(\frac{\alpha}{n}\right)\le n$

Then I did the same thing for $m\times \tan\left(\frac{\alpha}{m}\right)$ and reached the conclusion that:

$0\le m \tan \left(\frac{\alpha}{m}\right) \lt \infty$

Is this a valid solution for the problem? If not, I would appreciate it if someone could give me a hint to solve it.

• One approach: first show that $\frac{\sin(x)}{x}$ is decreasing and $\frac{\tan(x)}{x}$ is increasing on $[0,\pi/2)$. They coincide for $x \to 1$. Thus $\sin(x)/x \leq \tan(y)/y$ for all $x,y \in [0,\pi/2)$. Pick a suitable $x$ and $y$. – Winther Mar 24 '17 at 17:16
• Do you know that $\tan x>x$ and $\sin x<x$ for $x\in(0,\pi/2)$? – Thomas Andrews Mar 24 '17 at 17:24
• @Winther You mean for $x\to 0$, of course. – Thomas Andrews Mar 24 '17 at 17:29

Let $f(x)=-n \sin \frac{x}{n} +m\tan \frac{x}{m}$, where $0<x<\frac{\pi}{2}$
Hence, $f'(x)=\frac{1}{\cos^2\frac{x}{m}}-\cos\frac{x}{n}\geq0$.
Thus, $f(x)\geq f(0)=0$ and we are done!
Multiply both sides by $\alpha$ to express the inequality as $\dfrac{\sin x}x \le \dfrac{\tan y}{y}$ where $0 < x,y < \dfrac \pi 2$.
We apply the well known facts $\sin(x) \le x$ and $\tan(x) \ge x$ to find that $$n\sin(\alpha/n) \le \alpha$$ $$m\tan(\alpha/n) \ge \alpha$$ Equality occurs at $\alpha=0$