# Generalization of $f\left(\frac{x_1 + x_2}{2}\right) < {f(x_1) + f(x_2) \over 2}$ for odd number of variables.

Let $f: \mathbb R \to \mathbb R$ satisfy the following inequality:

$$f\left(\frac{x_1 + x_2}{2}\right) < {f(x_1) + f(x_2) \over 2}$$ Show that:

$$f\left(\frac{x_1 + x_2 + x_3}{3}\right) < {f(x_1) + f(x_2) + f(x_3) \over 3}$$

The above appeared to be fairly simple for the case of $x_1, \;x_2, \; x_3, \;x_4$. (Let's call that case $n=4$)

By the initial conditions we have: $$f\left(\frac{x_1 + x_2}{2}\right) < {f(x_1) + f(x_2) \over 2} \\ f\left(\frac{x_3 + x_4}{2}\right) < {f(x_3) + f(x_4) \over 2}$$

$$f\left(\frac{x_1 + x_2}{2}\right) + f\left(\frac{x_3 + x_4}{2}\right) < {f(x_1) + f(x_2) + f(x_3) + f(x_4) \over 2} \iff \\ \iff {1 \over 2}\left(f\left(\frac{x_1 + x_2}{2}\right) + f\left(\frac{x_3 + x_4}{2}\right)\right) < {f(x_1) + f(x_2) + f(x_3) + f(x_4) \over 4}$$

Let $p = \frac{x_1 + x_2}{2}$, $q = \frac{x_3 + x_4}{2}$, then using the initial conditions:

$$f\left(\frac{p + q}{2}\right) < \frac{f(p) + f(q)}{2}$$

And therefore:

$$f\left(\frac{x_1+x_2+x_3+x_4}{4}\right) < \frac{f(x_1) + f(x_2) + f(x_3) + f(x_4)}{4}$$

I see how this may be expanded further by induction for $n = 2^k$. However I failed to prove the case of $n = 3$.

How can I prove that the inequality holds for $n = 3$?

• Well, actually by induction you get the result for $n=2^k$. The idea for the other values of $n$ is to apply a 'down' version of induction. – Stan Tendijck Aug 7 '18 at 10:58
• @StanTendijck What is a 'down' version of induction? – roman Aug 7 '18 at 11:00
• With that I mean that given that it works for a general $n$ you could prove the $n-1$ case (by replacing the $n$th element with the mean of the first $n-1$) – Stan Tendijck Aug 7 '18 at 13:07

The inequality you are after is a special case of the proof presented in this answer. Let $\overline x=\frac{x_1+x_2+x_3}{3}$ and consider $$f(\overline x)=f\left(\frac{x_1+x_2+x_3+\overline x}{4}\right) <\frac{f(x_1)+f(x_2)+f(x_3)+f(\overline x)}{4}$$ where we have used the $n=4$ case you already have proven. Rearrange the inequality to get the desired result $$f(\overline x) <\frac{f(x_1)+f(x_2)+f(x_3)}{3}$$ We can in fact use the same technique as presented in the linked answer to prove a generalization, namely that $$f\left(\sum_{i=1}^n x_i t_i\right) < \sum_{i=1}^n t_i f(x_i)$$ for any rational $t_i$ with $\sum_{i=1}^n t_i = 1$.