How prove $\frac{\sin{a}-\sin{c}}{a-c}>\frac{\sin{b}-\sin{d}}{b-d}$

Question

Question:

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?

Check Rudin's real and complex analysis, page 61. This looks very similar. — Bombyx mori, May 30 '14 at 08:05
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. — Bombyx mori, May 30 '14 at 08:09

score 6 · Answer 1 · answered May 30 '14 at 08:20

6

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 $${\sin c-\sin a\over c-a}>{\sin c-\sin b\over c-b}>{\sin d-\sin b\over d-b}\ .$$

answered May 30 '14 at 08:20

Christian Blatter

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How prove $\frac{\sin{a}-\sin{c}}{a-c}>\frac{\sin{b}-\sin{d}}{b-d}$

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