How to prove the geometric mean is concave?
The geometric mean is $()=(\prod_{i=1}^nx_i)^{1/}$ is concave on dom =ℝ++.
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We are going to use the AM-GM inequality: $$\frac{a_1 + a_2+\cdots +a_n}{n} \ge (a_1a_2\ldots a_n)^{1/n}.$$
Applying AM-GM for $a_i = \frac{x_i}{\lambda x_i + (1 - \lambda)y_i}$, then for $a_i = \frac{y_i}{\lambda x_i + (1 - \lambda)y_i}$, we get:
\begin{align} \frac{f(x)}{f(\lambda x + (1 - \lambda)y)} = \left(\prod_{i=1}^n \frac{x_i}{\lambda x_i + (1 - \lambda)y_i} \right)^{1/n} &\le \frac{1}{n}\left(\sum_{i=1}^n \frac{x_i}{\lambda x_i + (1 - \lambda)y_i} \right),\\ \frac{f(y)}{f(\lambda x + (1 - \lambda)y)} =\left(\prod_{i=1}^n \frac{y_i}{\lambda x_i + (1 - \lambda)y_i} \right)^{1/n} &\le \frac{1}{n}\left(\sum_{i=1}^n \frac{y_i}{\lambda x_i + (1 - \lambda)y_i} \right). \end{align} Multiply the first inequality by $\lambda$, and the second by $(1 - \lambda)$, then sum up the two we get \begin{equation} \frac{\lambda f(x) + (1 - \lambda)f(y)}{f(\lambda x + (1 - \lambda)y)} \le 1. \end{equation} We are done.
f10w
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Maybe overkill but you could use Mahler's inequality to show this. Indeed,
$$ f(\lambda x + (1 - \lambda) y) = \prod_{i=1}^n (\lambda x_i + (1 - \lambda)y_i)^{1/n} \geq \prod_{i=1}^n \lambda^{1/n} x_i^{1/n} + \prod_{i=1}^n (1 - \lambda)^{1/n} y_i^{1/n} \\ = (\lambda^{1/n})^{n} \prod_{i=1}^n x_i^{1/n} + \left((1 - \lambda)^{1/n}\right)^n \prod_{i=1}^n y_i^{1/n} = \lambda f(x) + (1 - \lambda) f(y). $$
VHarisop
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For the third part it should be $(1 - \lambda)^{1/n} y_i^{1/n}$ rather than $(1 - \lambda)^{1/n} y_i^n$ – Frank Sep 25 '20 at 08:15
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@Frank good catch! Thanks. – VHarisop Sep 26 '20 at 05:41
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(+1) Not really overkill. f10w's answer essentially proves Mahler's inequality, so yours is essentially as simple. – robjohn Aug 22 '22 at 14:13