= Solution
Because all variables are positive, the product constraint is equivalent to $\sum_i\ln x_i=\ln a$. The <Lagrange multiplier> equations for
$$
\mathcal L=\sum_{i=1}^n x_i-\lambda\left(\sum_{i=1}^n\ln x_i-\ln a\right)
$$
are $1-\lambda/x_i=0$. Thus every $x_i$ equals $\lambda$, and the constraint requires $\lambda=a^{1/n}$. The candidate <minimum> is therefore
$$
\boxed{\min\sum_{i=1}^n x_i=na^{1/n},\qquad x_1=\cdots=x_n=a^{1/n}}.
$$
For a global justification, put $y_i=\ln x_i$. The objective becomes the strictly <convex function> $\sum_i e^{y_i}$ on the affine constraint $\sum_i y_i=\ln a$. A stationary point of this restricted <convex function> is its unique global <minimum>, establishing the claim without presupposing the desired inequality.
Finally take $a=\prod_i x_i$ and divide the minimum bound by $n$. This is the <arithmetic-geometric mean inequality>:
$$
\boxed{\frac1n\sum_{i=1}^n x_i\geq\left(\prod_{i=1}^n x_i\right)^{1/n}},
$$
with equality exactly when all the positive variables are equal.
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