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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?

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    You want to show that $\frac{\tan x}{x}$ is increasing. If you do it through the derivative, you want to show that $x-\cos x\sin x$ is positive. That can be done by comparing the area of a circular sector with angle $2x$ with the area of a certain triangle.2012-02-23

2 Answers 2

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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:

$x \gt \sin x, \forall x \in (0, \pi)$

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    thanks, that makes things simpler.2012-02-23
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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 $\int_a^u f(t)dt\ge \int_a^u g(t)dt$.