Let $a \in (0, \infty)$ be a D5407: Positive real number.

Then
\begin{equation}
N (1 / a, \ldots, 1 / a)
= \underset{x \in (0, \infty)^N : \sum_{n = 1}^N \frac{1}{x_n} = a}{\text{arg min }} \, \prod_{n = 1}^N x_n
\end{equation}

Result R5186
on D6039: Euclidean real component product operation

*Subresult of R5185: Tight lower bound to a finite product of positive real numbers*

Unique global minimizer for a finite product of positive real numbers with a given sum of multiplicative inverses

Formulation 0

Let $a \in (0, \infty)$ be a D5407: Positive real number.

Then
\begin{equation}
N (1 / a, \ldots, 1 / a)
= \underset{x \in (0, \infty)^N : \sum_{n = 1}^N \frac{1}{x_n} = a}{\text{arg min }} \, \prod_{n = 1}^N x_n
\end{equation}

Proofs

Let $a \in (0, \infty)$ be a D5407: Positive real number.

This result is a subresult to R5185: Tight lower bound to a finite product of positive real numbers. In that result, we see that $\prod_{n = 1}^N x_n$ attains the lowest possible value when $x_1 = x_2 = \cdots = x_N$. In this case, we have
\begin{equation}
\left( \frac{1}{\frac{1}{N} \sum_{n = 1}^N \frac{1}{x_n}} \right)^N
= \prod_{n = 1}^N x_n
= x^N_k
\end{equation}
for some $k = 1, \ldots, N$. Substituting $\sum_{n = 1}^N \frac{1}{x_n} = a$ and raising both sides to power $1 / N$, we have
\begin{equation}
\frac{N}{a}
= \frac{1}{a / N}
= x_k
\end{equation}
Since $k = 1, \ldots, N$ was arbitrary, the claim follows. $\square$