prove $\int_0^a\sin x\, dx + \int_0^b \arcsin x\, dx \geq ab$

Question

I want to prove that for

$$ 0\lt a \lt {\frac \pi 2}\\\\ 0\lt b \lt 1$$

then
$$\int_0^a \sin x\, dx + \int_0^b \arcsin x\, dx \geq ab$$

Thinking about these integrals geometric-wise, i thought that the integral on $\sin x$ is the area bounded between $y=0$ and $\sin x$ on the interval $[0,a]$, and the integral on $\arcsin x$ on the interval $[0,b]$ is in fact the area that completes the previous area to a rectangle, but in that case - i would get that $$\int_0^a \sin x\, dx + \int_0^b \arcsin x\, dx = ab$$

what do i miss here?

Answer 1

This is a special case of Laissant's inequality:

https://en.m.wikipedia.org/wiki/Integral_of_inverse_functions

From the drawing there you can see that there's a little rectangle which is not included in the two integrals. That is why the result is an inequality not an equality.

Answer 2

Your problem is a special case of the more general inequality below.

Young's Inequality

Let $f(x)$ be a continuous and strictly monotonous function, and its inverse function be $x=f^{-1}(y)$. Suppose that $f(0)=0$, $a>0$, $b>0$.Then $$\int_0^af(x){\rm d}x+\int_0^b f^{-1}(y){\rm d}y\geq ab,$$ with equality holding if and only if $f(a)=b.$

To prove it, you may apply the lemma below .

Lemma

Let $f(x)$ be continuous and strictly monotonous over $[a,b]$, and its inverse function be $x=f^{-1}(y)$. Suppose that $\alpha=f(a)$ and $\beta=f(b)$.Then $$\int_a^bf(x){\rm d}x+\int_\alpha ^ \beta f^{-1}(y){\rm d}y=b\beta-a\alpha.$$

If you want the detailed proofs, I'm glad to post them.