Einführung in die Theorie der Gammafunktion, Emil Artin, 1931 (The Gamma Function, trans. by Michael Butler, 1964, Holt, Rinehart, Winston) pp. 4–6, 7.

We shall call a function defined on an interval weakly convex if it satisfies the inequality $$f\left(\frac{x_1 + x_2}{2}\right) \le \frac{1}{2}(f(x_1) + f(x_2)) \;\; \textrm{(1.7)}$$ for all $$x_1$$, $$x_2$$ of the interval. It is obvious that the sum of two weakly convex functions, both defined on the same interval, is again weakly convex. It is also obvious that the limit function of a sequence of weakly convex functions, all defined on the same interval, is weakly convex.

Let $$f(x)$$ be weakly convex. The inequality (1.7) can be generalized to $$f\left(\frac{x_1 + \dots + x_n}{n}\right) \le \frac{1}{n}(f(x_1) + \dots + f(x_n)) \;\; \textrm{(1.8)}$$

Proof.

(1) We first show that if (1.8) holds for a certain integer $$n$$, then it also holds for $$2n$$. Indeed, suppose $$x_1, x_2, \dots, x_n$$ are numbers in our interval. Replacing $$x_1$$ and $$x_2$$ in Eq. (1.7) by $$\frac{x_1 + \dots + x_n}{n}$$ and $$\frac{x_{n+1}+\dots +x_{2n}}{n}$$, respectively, we have $$f\left(\frac{x_1+\dots +x_{2n}}{2n}\right) \le \frac{1}{2}\left(f\left(\frac{x_1+\dots +x_n}{n}\right) + f\left(\frac{x_{n+1}+\dots +x_{2n}}{n}\right)\right).$$ Applying the inequality (1.8) to both terms on the right-hand side, we get the desired formula $$f\left(\frac{x_1+\dots +x_{2n}}{2n}\right) \le \frac{1}{2n}(f(x_1) + \dots + f(x_{2n})).$$

(2) Next we show that if (1.8) holds for $$n + 1$$, then it also holds for $$n$$. With $$n$$ numbers $$(x_1, x_2, \dots, x_n)$$ the number $$x_{n+1} = \frac{1}{n}(x_1 + \dots + x_n)$$ also belongs to our interval. If (1.8) holds for $$n + 1$$, then $$f(x_{n+1}) = f\left(\frac{nx_{n+1} + x_{n+1}}{n+1}\right) = f\left(\frac{x_1 + \dots + x_n + x_{n+1}}{n+1}\right) \le \frac{1}{n+1}(f(x_1) + \dots + f(x_n) + f(x_{n+1})).$$ Transposing the term $$\frac{1}{n+1}f(x_{n+1})$$ to the left side, we obtain (1.8) for the $$n$$ given numbers.

(3) We now combine steps (1) and (2) to attain the desired result. If (1.8) holds for any integer $$n$$, then step (2) implies that it also holds for all smaller integers. Because of step (1) the contention is true for arbitrarily large integers. Therefore it must be true for all $$n$$. This completes the proof.

Numerous inequalities useful in analysis can be obtained from Eq. (1.8) by a suitable choice for $$f(x)$$. As an example, consider $$f(x) = - \log x$$ for $$x > 0$$. We have $$f''(x) = \frac{1}{x^2}$$ and our function is convex. Therefore Eq (1.8) implies that $$-\log \left(\frac{x_1 + \dots + x_n}{n}\right) \le -\frac{1}{n}(\log x_1 + \log x_2 + \dots + \log x_n) = -\log \sqrt[n]{x_1 x_2 \dots x_n}$$ and consequently $$\frac{x_1 + x_2 + \dots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \dots x_n}.$$