Prove $\sin^{-1}(1)\geq \int_0^b1/\sqrt{1-x^2}dx +(1-b)\pi/2$ for $b \in [0,1)$

Question

I'm trying to prove the following inequality:

$$\sin^{-1}(1)\geq\int_0^b1/\sqrt{1-x^2}\,dx +(1-b)\pi/2$$

for every $b \in [0,1)$.

I'm given $\sin^{-1}(1) = \pi/2$ and $\sin^{-1}(x)$ is strictly increasing. We also know $\sin^{-1}(x)$ is the inverse of the strictly increasing function $\sin(x)$ (when $x\in [-\pi/2, \pi/2] $).

My Attempt

I can prove using integration and the FTC that $\int_0^b1/\sqrt{1-x^2}\,dx = \sin^{-1}(b)$.

This information simplifies the inequality to $0 \geq \sin^{-1}(b) - b\times \pi/2$.

I'm having trouble showing that $ \sin^{-1}(b) \leq b\times \pi/2$ given that everything above is true.

Answer 1

The inequality you seek to prove is not true. Put $b=1/4$.

However, $\sin^{-1}b\le b\pi/2,\ \forall b\in[0,1)$

Let $f(x)=x\pi/2-\sin^{-1}x, f(0)=f(1)=0$

$f'(x)=\pi/2-\frac1{\sqrt{1-x^2}}$

$f'(x)>0,\ \forall x\in\Big[0,\sqrt{1-\frac4{\pi^2}}\Big)$ and $f'(x)<0,\ \forall x\in\Big(\sqrt{1-\frac4{\pi^2}},1\Big)$

This means that $f(x)$ initally increases from $f(0)$ to a maxima and then decreases to $f(1)$ in $[0,1]$. So $f(x)$ stays above $\min\{f(0),f(1)\}=0$. This means $f(x)\ge0\ \forall x\in[0,1)\implies f(b)\ge0$

Answer 2

I think $$\sin^{-1}(b) \leq b\frac{\pi}2$$ for every $b \in \big[0,1)$ since the graph of $y=b$ when scaled upto $y=\frac{\pi}2b\ $ will get above that of $\sin^{-1}(b)$ for $b\in\big(0,1)$ and equal for $b=0$.

Answer 3

The inequality you want to prove is $$ \arcsin1-\arcsin b<\frac{\pi}{2}(1-b). $$ This inequality is false for $b$ close to $1$. Indeed, by the Mean Value Theorem $$ \arcsin1-\arcsin b=\frac{1-b}{\sqrt{1-\xi^2}},\quad b<\xi<1. $$ If $$ \frac{1-b}{\sqrt{1-\xi^2}}<\frac{\pi}{2}(1-b), $$ then $$ b<\xi<\sqrt{1-\frac{4}{\pi^2}}. $$

Answer 4

If $f$ is strictly convex on $[0,1],$ then $(b,f(b))$ lies below the line through $(0,f(0))$ and $(1,f(1))$ for $b\in (0,1).$  This implies

$$\frac{f(1)-f(b)}{1-b} > \frac{f(1)-f(0)}{1}$$

for $b\in (0,1).$ Apply this with $f(b)=\arcsin b.$

Answer 5

Your inequality is reversed. The correct inequality is $$\arcsin(1)\geq \int_0^b\,\frac{1}{\sqrt{1-x^2}}\,\text{d}x+\frac{\pi}{2}\,(1-b)$$ for all $b\in[0,1]$. The equality holds if and only if $b=0$ or $b=1$.

As you did, we can show that the inequality above is equivalent to $$\arcsin(b)\leq \frac{\pi}{2}\,b$$ for all $b\in[0,1]$. Now, the function $f:=\arcsin$ is convex on $[0,1]$. Therefore, for each $b\in [0,1]$, $$f(b)=f\big((1-b)\cdot 0+b\cdot 1\big)\leq (1-b)\cdot f(0)+b\cdot f(1)$$ by Jensen's Inequality. This shows that $$\arcsin(b)\leq (1-b)\cdot 0+b\cdot\frac{\pi}{2}=\frac{\pi}{2}\,b\text{ for all }b\in[0,1]\,.$$ Since $\arcsin$ is strictly convex, the only equality cases are $b=0$ and $b=1$.