<s>[INST] How to prove this inequality $\sin{\left(\frac{\pi}{2}ab\right)}\le\sin{\left(\frac{\pi}{2}a\right )}\sin{\left(\frac{\pi}{2}b\right)}$? 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!
 [/INST] The proof is related to the concavity of the $\sin\left(\frac\pi2 x\right)$ curve.
Consider the following blue curve, which shows $y = \sin\left(\frac\pi2 x\right)$. The red dot $P$ is at $\left(A, \sin\left(\frac\pi2 A\right)\right)$. Now I fit (the red curve) a smaller version of $y = \sin\left(\frac\pi2 x\right)$ ending at $P$, so that a point $Q$ on the red curve split the width of the red rectangle to $B:(1-B)$ and the height to $\sin\left(\frac\pi2 B\right):\left[1-\sin\left(\frac\pi2 B\right)\right]$. The red curve is never below the blue curve. When we evaluate at $x=AB$, the green point $Q$ gives $\left(AB,\sin\left(\frac\pi2 A\right)\sin\left(\frac\pi2 B\right)\right)$ and the blue point $R$ gives $\left(AB,\sin\left(\frac\pi2 AB\right)\right)$.

</s>