TITLE: How to prove this inequality $\sin{\left(\frac{\pi}{2}ab\right)}\le\sin{\left(\frac{\pi}{2}a\right )}\sin{\left(\frac{\pi}{2}b\right)}$?
QUESTION [18 upvotes]: Let $$0\le a\le 1,0\le b\le 1$$

Prove or disprove
  $$\sin{\left(\dfrac{\pi}{2}ab\right)}\le\sin{\left(\dfrac{\pi}{2}a\right )}\sin{\left(\dfrac{\pi}{2}b\right)}$$

My try:
Since
$$\sin{x}\sin{y}=\dfrac{1}{2}[\cos{(x-y)}-\cos{(x+y)}]$$
So
$$\sin{\left(\dfrac{\pi}{2}a\right)}\sin{\left(\dfrac{\pi}{2}b\right)}=\dfrac{1}{2}[\cos{\dfrac{\pi}{2}(a-b)}-\cos{\dfrac{\pi}{2}(a+b)}]$$
Then I can't,Thank you very much!

REPLY [4 votes]: The stated inequality smells from "a convex curve on a doubly logarithmic graph paper". Therefore we consider the auxiliary function
$$f(t):=-\log\left(\sin\biggl({\pi\over2} e^{-t}\biggr)\right)\geq0\qquad(t\geq0)\ .$$
One computes
$$f'(t)={\pi\over2}\cot\left({\pi\over2}e^{-t}\right)\ e^{-t}>0,
\qquad f''(t)={\pi^2 e^{-2t}\over 4\sin^2\bigl({\pi\over2}e^{-t}\bigr)}\left(1-{\sin\bigl(\pi e^{-t}\bigr)\over \pi e^{-t}}\right)>0\ .$$
It follows that  $f'$ is increasing, whence we can write
$$f(t_1+t_2)=f(t_1)+\int_{t_1}^{t_1+t_2} f'(\tau)\ d\tau\geq f(t_1)+\int_0^{t_2}f'(\tau)\ d\tau=f(t_1)+f(t_2)\ .$$
Letting $a=e^{-t_1}$, $b=e^{-t_2}$ we now obtain
$$-\log\sin\left({\pi\over2} ab\right)\geq -\log\sin\left({\pi\over2} a\right)-\log\sin\left({\pi\over2} b\right)\ ,$$
which is equivalent to the stated inequality.