Prove that $x^2<\sin x \tan x$ as $x \to 0$ [duplicate]

Solution 1:

Apply GM - HM to $\sin x$ and $\tan x$ (both positive for $x\geq 0$), we get that

$$ \sqrt{ \sin x \tan x } \geq \frac{2} { \frac{1} {\sin x} + \frac{ 1}{ \tan x} } = \frac{2 \sin x} { 1 + \cos x } = 2 \tan \frac{x}{2} \geq x$$

The only 'calc' that you need is the last inequality, but that has an easy graphical approach too.

Solution 2:

The arctan substitution is a strictly monotone change of variable near the origin and therefore the string of inequalities are all equivalent to each other, so modulo checking the algebra the approach appears to be correct. Without necessarily knowing about Taylor series one could try to show that $\sin x \geq x - \frac{x^3}{6}$ while $\tan x > x + \frac{x^3}{3}$ from which the formula follows by algebra.