f : (a,b) → R is a midpoint-convex function (I didn't say continuity).

Here I'd like to verify following inequality ""directly"".

f( (x1+x2+x3)/3 ) ≤ (f(x1)+f(x2)+f(x3))/3


I can easily demonstrate for n=2^k

BUT not case for n=2^k, HOW can I demonstrate this inequallity?

Thanks for your consideration.


Assume you can do it for $n=2^k$, specifically $n=4$, then $f(\frac{(x_1+x_2+x_3+(x_1+x_2+x_3)/3)}{4})\leq \frac{f(x_1)+f(x_2)+f(x_3)+f((x_1+x_2+x_3)/3)}{4}$
rearrange to get n=3 case. Same approach can be generalized to any non $2^k$ number.

  • 1
    $\begingroup$ Thanks a lot. This is exactly what I want. (Sorry for being late check your answer.) $\endgroup$ – user143993 Sep 26 '14 at 6:28

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