Take $0<a<b<c<d$ and consider $f$ strictly increasing and strictly concave (with $f(0)\geq 0$). I would like to prove that $$\frac{f(d)-f(b)}{f(c)-f(a)}<\frac{d-b}{c-a}$$ without any further assumption on $a,b,c,d$. Is it possible? Thanks very much!
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Yes, this is true. See this answer. – Oct 13 '15 at 17:32
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Sorry Mice Elf, what is the 3-points convexity definition? – AlGri Oct 13 '15 at 17:44
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$f(ta+(1-t)b)\le tf(a)+(1-t)f(b)$ – Oct 13 '15 at 17:45
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The increasing assumption allows you to rearrange the desired inequality as $$\frac{f(d)-f(b)}{d-b}<\frac{f(c)-f(a)}{c-a}, \quad a<b<c<d \tag{1}$$ First, prove (1) with $b=c$, that is $$\frac{f(d)-f(b)}{d-b}<\frac{f(b)-f(a)}{b-a}, \quad a<b<d \tag{2}$$ This only requires writing out the strict concavity property $f(ta+(1-t)d)> tf(a)+(1-t)f(d)$ with a suitable $t$. (Such that $ta+(1-t)d = b$)
Once you have (2), proceed by applying it twice as follows: $$\frac{f(d)-f(b)}{d-b}<\frac{f(c)-f(b)}{c-b}<\frac{f(c)-f(a)}{c-a}$$