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!

Solution 1:

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 $b\in(0,1]$, $$f(\frac{\pi b}{2})\le f(\frac{\pi}{2}) \Longleftrightarrow \sin\frac{\pi ab}{2}\le \sin\frac{\pi a}{2}\sin\frac{\pi b}{2}.$$

Solution 2:

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.

Solution 3:

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 $\left(AB,\sin\left(\frac\pi2 AB\right)\right)$.