Prove that $\frac{\tan{x}}{\tan{y}}>\frac{x}{y} : \forall (0Prove that $\frac{\tan{x}}{\tan{y}}>\frac{x}{y} : \forall (0<y<x<\frac{\pi}{2})$.
My try, considering $f(t)=\frac{\tan{x}}{\tan{y}}-\frac{x}{y}$ and derivating it to see whether the function is increasing in the given interval.
I should be sure that $\lim_{x,y\rightarrow0}\frac{\tan{x}}{\tan{y}}-\frac{x}{y}\geq0$ for the previous derivative check to be useful, which I'm not yet, but I'm assuming it's $0$ since I'd say that since both $x,y$ approach $0$ equally then the quotient of both their tangents and themselves is $1$, hence the substraction being $0$.
However, the trouble arrives at the time of derivating it because of the 2 variables, I'm not sure if I have to fix one and derivate in terms of the other one, or what to do. I have to say I'm currently coursing a module on real single-variable analysis, so it can't have anything to do with multivariable analysis.
 A: Consider the function
$$f(x) = \frac{\tan x}{x}.$$
You need to show that function is increasing on $(0,\pi/2)$.
If you differentiate that, you find
$$f'(x) = \frac{x(1+\tan^2 x) - \tan x}{x^2},$$
and you need to show $x(1+\tan^2 x) > \tan x$ for $x\in (0,\pi/2)$. A little trick helps showing that.
A: In fact, we can show a slightly stronger result.

If $f(0)=0$ and $f(x)$ is strictly convex then for $x<y$ we have $$\frac{f(x)}x<\frac{f(y)}y.$$

This result is shown here for differentiable functions. But it can be shown for any convex function.
Proof. Since $x=\frac xy \cdot y + \frac{y-x}y \cdot 0$ we get
$$f(x) < \frac xy f(y) + \frac{y-x}y f(0)$$
from convexity. Hence
$$\frac{f(x)}x < \frac{f(y)}y.$$

So for $f(x)=\tan x$ we have
$$\frac{\tan x}x < \frac{\tan y}y$$
whenever $0<x<y<\frac\pi2$. This is equivalent to the inequality in the original question.
A: The inequality is equivalent to
$$
\frac{\cot y}{\cot x}>\frac{x}{y}
$$
or
$$
y\cot y>x\cot x
$$
so one could try proving that $f(x)=x\cot x$ is decreasing in $(0,\pi/2)$. Now
$$
f'(x)=\cot x-\frac{x}{\sin^2x}=\frac{\sin x\cos x-x}{\sin^2 x}=\cdots
$$
(a simple transformation of the numerator will do).
A: If we can show that $\dfrac{\tan(x)}{x}$ is monotonically increasing, then if $0\lt y\lt x\lt\dfrac\pi2$, we have
$$
\frac{\tan(x)}{x}\gt\frac{\tan(y)}{y}\quad\text{and therefore}\quad\frac{\tan(x)}{\tan(y)}\gt\frac xy
$$
The derivative of $\dfrac{\tan(x)}{x}$ is
$$
\frac{x\sec^2(x)-\tan(x)}{x^2}=\frac{\sec^2(x)}{2x^2}\left(2x-\sin(2x)\right)
$$
Now all you need to show is that $2x-\sin(2x)\gt0$ on $(0,\frac\pi2)$.
A: The function $x \cot x$ is strictly decreasing on $[0,\pi)$, as in other answers.
It turns out that the Taylor expansion at $0$ is
$$x \cot x = 1 - \frac{x^2}{3}- \frac{x^4}{45} - \frac{2 x^6}{945} - \cdots $$
valid on $(-\pi, \pi)$, check the general formula here. Notice that all of the coefficients after the free term are $\le 0$.
