Let n be a positive integer.
What is the geometric mean of the divisors of n?

Let the divisors of n be d_{1}, d_{2}, ..., d_{k}.
The geometric mean of the product of the divisors is
(d_{1}d_{2}...d_{k})^{1/k}.

If n is not a square number, then the divisors can be organized into
pairs whose product is n
(e.g. the divisors of 12 are 1 and 12, 2 and 6, 3 and 4).
In this case, k is an even number, so let k = 2m.
Ordering the divisors into pairs, we can write the geometric mean as:

(d_{1}d_{2})^{1/k}(d_{3}d_{4})^{1/k}...(d_{m-1}d_{m})^{1/k}

= n^{1/k}n^{1/k}...n^{1/k}

= n^{m/k}

= n^{1/2}.

If n is a square, then it its square root as one divisor, and all the
other divisors come in pairs as before.
Write k = 2m + 1, and the geometric mean is:

(d_{1}d_{2})^{1/k}(d_{3}d_{4})^{1/k}...(d_{m-1}d_{m})^{1/k}n^{1/2k}

= n^{1/k}n^{1/k}...n^{1/k}n^{1/2k}

= n^{m/k}n^{1/2k}

= n^{(2m+1)/2k}

= n^{1/2}

In all cases, then, the geometric mean of the divisors of n is
equals to the square root of n.