simple Inequality.

Question

Prove: if $x \in (0,1)$ then: $$ \sqrt{\frac{1-x}{1+x}} < \frac{\ln{(1+x)}}{\sin^{-1}x} < 1$$

I have been thinking on using maximization to solve this problem, but the function is not nice! Is there a easy way to solve this problem? Any trick?

Answer 1

Let $\sin^{-1}(x) = \theta \in (0, \pi/2)$. First let us prove that $\dfrac{\log(1+\sin(\theta))}{\theta} < 1$.

Consider $f(\theta) = \theta - \log(1+\sin(\theta))$. $$f'(\theta) = 1 - \dfrac{\cos(\theta)}{1+\sin(\theta)} > 0$$ since $\cos(\theta) < 1 < 1 + \sin(\theta)$ for $\theta \in (0, \pi/2)$. Hence $f(\theta)$ is increasing in $(0, \pi/2)$. Hence, $f(\theta) > f(0)$.

To prove the next inequality i.e. $$\sqrt{\dfrac{1-\sin(\theta)}{1+\sin(\theta)}} < \dfrac{\log(1+\sin(\theta))}{\theta}$$ Note that $\sqrt{\dfrac{1-\sin(\theta)}{1+\sin(\theta)}} = \dfrac{\cos(\theta/2) - \sin(\theta/2)}{\cos(\theta/2) + \sin(\theta/2)}$ for $\theta \in (0, \pi/2)$.

Now consider the function $g(\theta) = (1+\tan(\theta/2)) \log(1+\sin(\theta)) - \theta(1-\tan(\theta/2))$ and prove that it is increasing.

Answer 2

Hint - prove first the inequalities (for $x\in(0,1)$):

$$ x < \arcsin x < \frac{\pi}{2}x,$$ $$ \frac{x}{x+1} < \log(1+x) < x, $$

then use them in order to prove two simpler, algebraic inequalities.