let $a,b,c,d$ such $0<a<b<c<d<\pi$, show that $$\dfrac{\sin{a}-\sin{c}}{a-c}>\dfrac{\sin{b}-\sin{d}}{b-d}$$

My idea: if we use Mean value theorem then there exsit $$\xi\in(a,c),\eta\in(b,d)$$ such $$\cos{\xi}=\dfrac{\sin{a}-\sin{c}}{a-c},\cos{\eta}=\dfrac{\sin{b}-\sin{d}}{b-d}$$ but for $\cos{\xi}$ and $\cos{\eta}$, which is greater? we cannot know, because maybe we have $\xi=\eta$.

So how to prove this inequality?

  • $\begingroup$ This is Jensen's inequality, isn't it? $\endgroup$ – Bombyx mori May 30 '14 at 8:03
  • $\begingroup$ @Bombyxmori,I fell is not $\endgroup$ – math110 May 30 '14 at 8:04
  • $\begingroup$ Check Rudin's real and complex analysis, page 61. This looks very similar. $\endgroup$ – Bombyx mori May 30 '14 at 8:05
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    $\begingroup$ I think this is just the geometric form of Jensen's inequality for $\sin[x]$, the only difference is you have to shift $b$ equal to $c$ to change it into Rudin's form. But this should not be difficult to prove. $\endgroup$ – Bombyx mori May 30 '14 at 8:09

Draw a figure. Looking at the slopes of various segments you then can immediately verify that the concavity of $\sin$ in the interval $[0,\pi]$ implies $${\sin c-\sin a\over c-a}>{\sin c-\sin b\over c-b}>{\sin d-\sin b\over d-b}\ .$$


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