Proposition 2 ("geometric mean inequality")

Tag 089T, permanent. Cite it as [Bourbaki, Tag 089T].

Functions of a Real Variable, chapter III, § 1. DERIVATIVES OF THE EXPONENTIAL AND CIRCULAR FUNCTIONS

For any numbers $z_i > 0$ ( $1 \leq i \leq n$ ) and numbers $p_i > 0$ such that $\sum_{i=1}^n p_i = 1$ , one has

z_1^{p_1} z_2^{p_2} \cdots z_n^{p_n} \leq p_1 z_1 + p_2 z_2 + \cdots + p_n z_n.

(6) Moreover, the two sides of (6) are equal only if the

z_i

are equal.

Let us put $z_i = e^{x_i}$ ; then the inequality (6) can be written

\exp(p_1 x_1 + p_2 x_2 + \cdots + p_n x_n) \leq p_1 e^{x_1} + p_2 e^{x_2} + \cdots + p_n e^{x_n}.

(7) The proposition thus follows from prop. 1 of I, p. 26 applied to the function

e^x

, which is strictly convex on

\mathbf{R}

One says that the left- (resp. right-) hand side of (6) is the weighted geometric mean (resp. weighted arithmetic mean) of the $n$ numbers $z_i$ relative to the weights $p_i$ ( $1 \leq i \leq n$ ). If $p_i = 1/n$ for $1 \leq i \leq n$ , one calls the corresponding arithmetic and geometric means the ordinary arithmetic and geometric means of the $z_i$ . Then the inequality (6) can be written

(z_1 z_2 \cdots z_n)^{1/n} \leq \frac{1}{n}(z_1 + z_2 + \cdots + z_n).

(8)

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Nothing.

Cites

I, p. 26