Proof via Jensen's inequality (concavity of ln)
This proof assumes $x_1,\dots,x_n>0$ (strictly positive). For nonnegative numbers the statement follows by continuity or by a simple separate check when some entries are zero.
Statement. For positive real numbers $x_1,\dots,x_n$, \[\frac{x_1+\cdots+x_n}{n} \ge \bigl(x_1\cdots x_n\bigr)^{1/n}. \]
Proof. Consider the function $f(t)=\ln t$, which is concave on $(0,\infty)$ since $f''(t) = -1/t^2 < 0$. By Jensen's inequality for the equal weights $1/n$ we have \[f\Bigl(\frac{x_1+\cdots+x_n}{n}\Bigr) \ge \frac{1}{n}\sum_{i=1}^n f(x_i) = \frac{1}{n}\sum_{i=1}^n \ln x_i. \] Exponentiating both sides yields \[\ln\Bigl(\frac{x_1+\cdots+x_n}{n}\Bigr) \ge \frac{1}{n}\sum_{i=1}^n \ln x_i \quad\Longrightarrow\quad \frac{x_1+\cdots+x_n}{n} \ge \exp\Bigl(\frac{1}{n}\sum_{i=1}^n \ln x_i\Bigr) = \bigl(x_1\cdots x_n\bigr)^{1/n}. \] Equality holds iff all $x_i$ are equal (the equality condition for Jensen on a strictly concave function).
Remark. If some $x_i=0$ then the geometric mean is $0$ and the inequality is trivial since the arithmetic mean is nonnegative. Thus the result holds for all nonnegative $x_i$.