Does anyone know how to prove that if $a<b<c<d$, then $\frac{f(d)-f(b)}{d-b}\ge \frac{f(c)-f(a)}{c-a}$ holds. It is obvious to conclude by drawing a graph, but I have no idea how to prove mathematically.
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1I may know how to do now. We can easily show that $g(x,y):=\frac{f(x)-f(y)}{x-y}$ is an increasing function both in $x$ and $y$ by the convexity of $f$. Then we can show that that inequality holds. Is it okay? – Pin Nov 12 '17 at 23:30
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Perfectly fine. – Jack D'Aurizio Nov 13 '17 at 13:03
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Hint: Use the following lemma: $$\dfrac{f(x)-f(y)}{x-y}\leq\dfrac{f(x)-f(z)}{x-z}+\dfrac{f(y)-f(z)}{y-z}$$ for $x<y<z$ and $f$ convex.
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