I am trying to prove the following inequality I came across while looking at a geometric problem: $$\frac{(b-a)\sin{a}}{\sin{(b-a)}}\leq 1$$ with the constraints $0\leq a<b \leq \frac{\pi}{2}$.

I have confirmed this using software but haven't been able to prove it. Any help will be appreciated

  • $\begingroup$ what's the geometric problem you came across $\endgroup$ – Oliver Diaz Jan 30 at 2:01

Define $ u = b-a \in [0,\frac{\pi}{2}]$.

Rewrite the claim as $u \sin (a) \leq \sin(u)$. We fix $a$ and study $\frac{\sin(u)}{u}$ when $u \in [0, \frac{\pi}{2}-a]$.

If you differentiate $\frac{\sin(u)}{u}$, you will find no turning points on $[0,\frac{\pi}{2}]$. Thus since $\frac{\sin(u)}{u}$ is decreasing,

$\frac{\sin(u)}{u} \geq \frac{\sin(\frac{\pi}{2}-a)}{\frac{\pi}{2}-a}$.

Define $c = \frac{\pi}{2}-a$. We have

$ \frac{\sin(c)}{c}$.

Our right hand side is $\sin(a) = \cos(c)$.

Thus the claim follows since $\sin(c) \geq c \cos(c)$ on $[0,\frac{\pi}{2}]$.

  • $\begingroup$ Nice approach, thank you $\endgroup$ – Matheo Jan 30 at 2:48
  • $\begingroup$ Your post seems unclear from when you introduce $c$ and onward. Could you elaborate? $\endgroup$ – WaveX Jan 30 at 3:28
  • $\begingroup$ note that $\sin(c) = \cos(a)$, $\cos(c) = \sin(a)$. We want to prove that $\frac{\sin(\frac{\pi}{2}-a)}{\frac{\pi}{2}-a} \geq \sin(a)$. This is equivalent to proving $\frac{\sin(c)}{c} \geq \sin(a) = \cos(c)$. $\endgroup$ – fGDu94 Jan 30 at 3:32
  • $\begingroup$ I understand all of that. What let's us say the last inequality? $\endgroup$ – WaveX Jan 30 at 3:40
  • $\begingroup$ It's a standard one. You can prove it by differentiating $\sin(c)-c\cos(c)$ and proving it is positive on this interval. Notice that the inequality holds at $c=0$. $\endgroup$ – fGDu94 Jan 30 at 4:26

I hope this helps. On $[0,\pi/2]$ the function $t\mapsto \frac{\sin t}{t}$ is decreasing, and $t\mapsto \sin t$ is increasing. So, if $0\leq a<b\leq\frac{\pi}{2}$,

$$\frac{\sin a}{b}\leq \frac{\sin b}{b}\leq \frac{\sin(b-a)}{b-a}$$

Hence $$ \frac{(b-a)\sin a}{\sin(b-a)}\leq b $$

The inequality you suggested follows for $b\leq 1$. For the other cases it still escapes me.

  • $\begingroup$ Nice idea, thanks for the help. $\endgroup$ – Matheo Jan 30 at 2:49

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