Prove $\sin^{-1}(1)\geq \int_0^b1/\sqrt{1-x^2}dx +(1-b)\pi/2$ for $b \in [0,1)$ 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. 
 A: 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
$$
b<\xi<\sqrt{1-\frac{4}{\pi^2}}.
$$
A: 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 $f(x)\ge0\ \forall x\in[0,1)\implies f(b)\ge0$
A: 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 $b=0$.
A: 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 $f(b)=\arcsin b.$
A: 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 $b=1$.
