How can it be proved that the geometric mean function is concave?

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?

There may be a clever way to prove concavity without the Hessian, but I don't see one. So, here is the Hessian (I'm working under assumption $x_i>0$ for all $i$): $$D_{ij}f=\frac{f}{n^2}A \quad \text{where } \ A_{ij}= \begin{cases}(1-n)x_i^{-2} \quad &\text{ if }\ i=j \\ x_i^{-1}x_j^{-1} \quad &\text{ if }\ i\ne j \end{cases} \tag1 $$ Let $y_i=1/x_i$ to simplify notation. We are to prove that $v^TAv\le 0$ for every vector $v$. And indeed, $$v^TAv=\left(\sum_{i=1}^n y_iv_i\right)^2-n \sum_{i=1}^n y_i^2 v_i^2 \le 0\tag2$$ by the Cauchy-Schwarz inequality applied to $1\cdot (y_iv_i) $.

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.