I bumped into this question:
Question: Prove that for $x\in \Bigl[0,\dfrac {\pi}{4}\Bigr]$, $$\sin (\tan x) \geq {x}$$
This seems to be an innocent inequality but I am already exhausted trying to prove it. I followed different lines of thought and mention $3$ of them below:
$1.$ I tried using $\sin x<x<\tan x$. Knowing that $\sin x$,$\tan x$ increase in the mentioned interval, I tried applying $\tan$ and $\sin$ on different sides of $\sin x<x<\tan x$ in order to achieve something like that in the question.. But I just could not come to the inequality in the question.
$2.$ I tried using the function $f(x)=\sin(\tan x)-x$. $f(0)=0$. If we prove that the derivative $f'(x)=(\sec ^2 x)(\cos(\tan x))-1$ is positive, then we will be done. But proving even this seems very difficult.
$3.$ Since $\arcsin x$ is one-one and increasing in the mentioned interval, we can take $\arcsin$ on both sides of the inequality $\sin (\tan x) \geq {x}$ to get $\tan x > \arcsin x$. Thus proving $\tan x > \arcsin x$ would be equivalent to solving the problem.
All the three approaches are seeming impossible.. Any suggestions or other methods?
Thanks..
Best Answer
Let $t = \tan x$. Then you need to prove $\sin t \geq \arctan t$ for $t \in [0,1]$.
If you look at $f(t) = \sin t - \arctan t$, we have $f'(t) = \cos t - \frac{1}{1 + t^2}$. But the inequalities $$\frac{1}{1 + t^2} \leq 1 - \frac{1}{2}t^2 \leq \cos t, \quad (0 \leq t \leq 1)$$ are straightforward, showing that $f'(t) \geq 0$. Therefore $f$ is increasing on $[0,1]$. Because $f(0) = 0$, it follows that $f(t) \geq 0$, which is the desired inequality.
To prove the inequality $\cos t \geq 1 - \frac{1}{2}t^2$, let $g(t) = \cos t - 1 + \frac{1}{2}t^2$, and show that $g(0) = 0$, $g'(t) \geq 0$ for $t \geq 0$.