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

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) $.