Showing $\frac{\tan x}{x} \leq \frac{\tan y}{y}$ for all $x,y$ such that $0 \lt x \leq y < \pi/2$

Question

I was reviewing some old calculus notes and came across this question:

Show that the inequality $$\frac{\tan x}{x} \leq \frac{\tan y}{y}$$ is valid for every $x,y$ such that $0 \lt x \leq y < \pi/2$

This one stumped me a bit. Does anyone have any idea about how I can approach this without using graphs?

Much appreciated!

See for example https://math.stackexchange.com/q/631918 or https://math.stackexchange.com/q/719814 or https://math.stackexchange.com/q/724012 — Martin R, May 16 '22 at 09:33
$\sin x \cos x \leq (x) (1)$. Use this to show that derivative of $\frac {\tan x} x$ is non-negative. — Kavi Rama Murthy, May 16 '22 at 09:35
Does this answer your question? Prove that $\frac{\tan{x}}{\tan{y}}>\frac{x}{y} : \forall (0<y<x<\frac{\pi}{2})$ — Martin R, May 16 '22 at 09:37
This is a property of convex functions, see https://math.stackexchange.com/a/1834388. — Martin R, May 16 '22 at 09:39

score -1 · Answer 1 · answered May 16 '22 at 10:10

-1

The function $t \mapsto f(t)=\dfrac{\tan t}{t}$ in increasing in the interval $(0,\frac{\pi}{2})$ so, when $y\ge x$, we have that $f(y) \ge f(x)$.

answered May 16 '22 at 10:10

PierreCarre

1 Answers1