First of all, let’s recall that a convex function on a open
interval (a,b) is continuous
on (a,b) and admits left and right
derivative f+(x) and f-(x) for any x∈(a,b). For this reason,
it’s always possible to construct at least one supporting line (http://planetmath.org/ConvexFunctionsLieAboveTheirSupportingLines) for f(x) at any x0∈(a,b) : if f(x0) is
differentiable

in x0, one has r(x)=f(x0)+f′(x0)(x-x0); if not, it’s obvious that all r(x)=f(x0)+c(x-x0) are supporting lines for
any c∈[f-(x0),f+(x0)].
Let now r(x)=f(a+b2)+c(x-a+b2) be a supporting line of f(x) in x=a+b2∈(a,b).
Then, r(x)≤f(x). On the other side, by
convexity definition, having defined s(x)=f(a)+f(b)-f(a)b-a(x-a) the line connecting the points (a,f(a)) and (b,f(b)) , one has f(x)≤s(x). Shortly,
Integrating both inequalities
between a and b
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∫bar(x)𝑑x≤∫baf(x)𝑑x≤∫bas(x)𝑑x |
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∫bar(x)𝑑x |
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= |
∫ba[f(a+b2)+c(x-a+b2)]𝑑x |
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= |
f(a+b2)(b-a)+c∫ba(x-a+b2)𝑑x |
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= |
f(a+b2)(b-a) |
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∫bas(x)𝑑x |
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= |
∫ba[f(a)+f(b)-f(a)b-a(x-a)]𝑑x |
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= |
f(a)(b-a)+f(b)-f(a)b-a∫ba(x-a)𝑑x |
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= |
f(a)+f(b)2(b-a) |
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and so
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f(a+b2)(b-a)≤∫baf(x)𝑑x≤f(a)+f(b)2(b-a) |
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which is the thesis.