Difference between averaging and taking n-th root of products

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It seems like taking the n-th root of the product of n values is analogous to the traditional form of averaging. For example:

$$ \frac{5+5+5}{3} = 5 = \sqrt[\leftroot{0}\uproot{0}3]{(5*5*5)} $$

But they do not always produce the same results

$$ \frac{1+2+3}{3} = 2 $$ $$ \sqrt[\leftroot{0}\uproot{0}3]{(1*2*3)} = 1.817... $$

Also, the n-th root "average" cannot handle negative numbers. Anyway, what are the benefits of this latter form of averaging versus the standard way?

(Context: I'm in a course where "perplexity" was intuitively defined as the "average amount of surprise". It made me wonder why we don't just use the standard way of averaging.)

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The geometric mean can be thought of as $e^y$ where $y$ is the average of the natural logarithms of your values. That is $$\left(\prod_{i=1}^n x_i\right)^\frac 1 n = e^{ {\left( \sum_{i=1}^n \ln x_i \right)} \over n}$$Thus this sort of average is useful when your values can be thought of as powers of some base, and you want to find the average value of the exponent.