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!

The statement is true. 
Proof: Given $a\in[0,1]$, define
$$f(x)=\frac{\sin( ax)}{\sin x},\quad x\in (0,\frac{\pi}{2}],\quad\text{and}\quad g(x)=a\tan x-\tan(ax),\quad  x\in[0,\frac{\pi}{2}).$$
From $g(0)=0$ and
$$g'(x)=a \big((\sec x)^2-(\sec (ax))^2\big)\ge 0,\quad\forall x\in (0,\frac{\pi}{2}),$$
we know that $g(x)\ge 0$ on $(0,\frac{\pi}{2})$. Therefore,
$$f'(x)=\frac{a\sin x\cdot \cos(ax)-\cos x\cdot\sin(ax)}{(\sin x)^2}=\frac{\cos x\cdot\cos(ax)}{(\sin x)^2}\cdot g(x)\ge 0,\quad\forall x\in (0,\frac{\pi}{2}).$$
It follows that $f(x)\le f(\frac{\pi}{2})$ on $(0,\frac{\pi}{2}]$. 
Now we can complete the proof. When $b=0$, the inequality is trivial; when  $b\in(0,1]$,
$$f(\frac{\pi b}{2})\le f(\frac{\pi}{2}) \Longleftrightarrow \sin\frac{\pi ab}{2}\le  \sin\frac{\pi a}{2}\sin\frac{\pi b}{2}.$$
