Prove the following:
1) $∀x \in [0,1]$: $ x ≤ \arcsin(x) $≤ $\frac{x}{\sqrt{1−x^2}}$
2) $∀a,b \in {\displaystyle \mathbb {R_+} }$ : $a < b$ $\Rightarrow$ $\frac{b-a}{b} < log\left(\frac ba\right) < \frac {b-a}a $
I thought about it and still don't know where to start.
Any ideas? Thanks!
Hint : Let $x =\sin t$.
So you need to prove that
$$\sin t \le t \le \tan t$$
Can you do that? (Using derivatives)
The integral representation of the arcsine function is given by
$$\arcsin(x)=\int_0^x \frac{1}{\sqrt{1-t^2}}\,dt \tag 1$$
If $t\in [0,x]$, and $1>x\ge 0$, then we have
$$\sqrt{1-x^2}\le\sqrt{1-t^2}\le 1\tag 2$$
Using $(2)$ in $(1)$ we see that for $1>x\ge 0$
$$x\le \arcsin(x)\le \frac{x}{\sqrt{1-x^2}}$$
The integral representation of the logarithm function is given by
$$\log(x) =\int_1^x \frac{1}{t}\,dt$$
for $x>0$.
Hence, we have for $0<a<b$
$$\log(b/a)=\int_a^b \frac1t\,dt \tag 3$$
Clearly, for $0<a<t<b$, we have form $(3)$
$$\frac{b-a}{b}<\log(b/a)<\frac{b-a}{a}$$
PRE-CALCULUS APPROACH:
I though it might be instructive to present an approach that relies on elementary, pre-calculus tools only. To that end, we proceed.
In THIS ANSWER, I used only the limit definition of the exponential function and Bernoulli's Inequality to show that the logarithm function satisfies the inequalities
$$\frac{x-1}{x}\le \log(x)\le x-1 \tag4$$
Letting $x=b/a$ yields the coveted result
$$\frac{b-a}{b}\le \log(b/a)\le \frac{b-a}{a}$$
And in THIS ANSWER, I used the well-known inequalities
$$x\le \sin(x)\le x\cos(x)$$
for $x\in [-\pi/2,0]$ from elementary geometry to show that the arcsine function satisfies the inequalities
$$\frac{x}{\sqrt{1-x^2}} \le \arcsin(x)\le x$$
for $x\in (-1,0)$. Exploiting the odd symmetry we see immediately that for $x\in (0,1)$ satisfies the inequalities
$$x\le \arcsin(x)\le \frac{x}{\sqrt{1-x^2}}$$