TITLE: How can it be proved that the geometric mean function is concave?
QUESTION [9 upvotes]: A function $f: \mathbb R ^n \rightarrow \mathbb R $  is said to be concave if 
$\forall x,y \in \mathbb{R}^n, \forall \lambda \in [0,1]$ we have $ \lambda f(x) + (1-\lambda)f(y) \le f( \lambda x + (1- \lambda)y)$. 
In the case of the arithmetic mean function $f(x_1, ...,x_n) = (x_1 ...x_n)^{1/n}$ how would we prove convexity? I have been trying all day to find a proof, mostly by induction, but also considering the Hessian, which if always negative semidefinite implies convexity. Any tips?

REPLY [3 votes]: I was suggested this question while answering a duplicate. I'm cross-posting my proof here for the sake of future readers:
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.