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PostPosted: Sun, 20 Feb 2011 15:40:02 UTC 
S.O.S. Oldtimer

Joined: Sun, 4 Nov 2007 12:08:30 UTC
Posts: 246
Location: Bratislava, Slovakia
Probably too easy for experienced members, but at least for someone this might be a good exercise to get acquainted with a technique.

Let f: R\to R be a midpoint convex function, i.e.
$f\left(\frac{x+y}2\right) \le \frac{f(x)+f(y)}2$
for any x,y \in R.
Show that then this function fulfills
$f(t x + (1-t)y)\le t f(x) + (1-t)f(y)$
for any x,y\in R and any rational number t\in\langle0,1\rangle.

Cauchy induction: wikipedia

Sketch of the solution:
It is relatively easy to see, that it suffices to show f([x_1+\dots+x_k]/k)\le [f(x_1)+\dots+f(x_k)]/k for any integer k (and any choice of x_1,\dots,x_k\in R).

The case k=2^n is a straightforward induction.

Now, if 2^{n-1}<k\le 2^n then we denote \overline x=\frac{x_1+\dots+x_k}k. Now from
$f(\overline x)=f\left(\frac{x_1+\dots+x_k+\overline x+\dots+\overline x}{2^n}\right) \le\frac{f(x_1)+\dots+f(x_k)+(2^n-k)f(\overline x)}{2^n}$
we get kf(\overline x) \le f(x_1+\dots+f(x_k) by a simple algebraic manipulation.

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