TITLE: Prove that for every $0<\alpha<\beta<\pi/2$: $\tan\beta/\tan\alpha>\beta/\alpha$
QUESTION [6 upvotes]: Prove that for every $0<\alpha<\beta<(\pi/2)$:
 $$\displaystyle\frac{\tan\beta}{\tan\alpha}\gt\frac{\beta}{\alpha}$$ 

I tried setting a function $f(x) = \tan(x)$ and using the Mean Value theorem to prove this, but that didn't work since $f(0)$ is undefined. 
Does this involve some sort of trig identity?

REPLY [7 votes]: Hint: 
Consider $$f(x) = \frac{\tan x}{x}, x \in (0,\frac{\pi}{2})$$
and try to investigate if it is increasing or decreasing.
Also, try and prove that:
$$x \gt \sin x, \forall x \in (0, \pi)$$

REPLY [7 votes]: The inequality is equivalent to
$$\frac{\tan\beta}{\beta} >\frac{\tan\alpha}{\alpha}.$$
This says that $(\tan x)/x$ is an increasing function on $(0,\pi/2)$. Differentiate:
$$\frac{d}{dx}\frac{\tan x}{x}=\frac{2x-\sin2x}{2}\left(\frac{\sec x}{x}\right)^2.$$
For $(\tan x)/x$ to be increasing we want to check that the above is be positive on the interval, which after dividing by obviously positive terms we deduce is equivalent to $u>\sin u$ on $u\in(0,\pi)$. We can deduce this by noting that both $u'=1=\sin'(u)$ at $u=0$, and $u'=1>\sin'u=\cos u$ on $[0,\pi)$. This implies $u>\sin u$ because, in general, $f\ge g$ on an interval entails $\int_a^u f(t)dt\ge \int_a^u g(t)dt$.