Is there a geometric analog of absolute value?

I'm wondering whether there exists a geometric analog concept of absolute value. In other words, if absolute value can be defined as

$$ \text{abs}(x) =\max(x,-x) $$

intuitively the additive distance from $0$ to $x$, is there a geometric version

$$ \text{Geoabs}(x) = \max(x, 1/x) $$

which is intuitively the multiplicative "distance" from $1$ to $x$?

Update: Agreed it only makes sense for $Geoabs()$ to be restricted to positive reals.

To give some context on application, I am working on the solution of an optimization problem something like:

$$ \begin{array}{ll} \text{minimize} & \prod_i Geoabs(x_i) \\ \text{subject to} & \prod_{i \in S_j} x_i = C_j && \forall j \\ &x_i > 0 && \forall i . \end{array} $$

Basically want to satisfy all these product equations $j$ by moving $x_i$'s as little as possible from $1$. Note by the construction there are always infinite feasible solutions.

To make things easier I'll set $f(x)=\max\{x,-x\}$ and $g(x)=\max\{x,\frac{1}{x}\}$.

So we understand that $f:\mathbb{R}\to \mathbb{R}^+$ and $g: \mathbb{R}^+\to \mathbb{R}^+$.

Then $\exp(f(x))=g(\exp(x))$. So we can use this to translate some properties like the triangle inequality.

$$ g(xy)=g(\exp(\log(xy)))=\exp(f(\log(xy)))=\exp(f(\log(x)+\log(y))) $$ $$ \leq \exp(f(\log x)+f(\log y))=\exp(f(\log x))\exp(f(\log y))=g(\exp(\log(x))g(\exp(\log(x)) $$ $$ =g(x)g(y) $$

So $g(xy)\leq g(x)g(y)$ and we have the multiplicative triangle inequality.

Of course this is easier to show directly but the method emphasizes the "transfer".

Another good sign is $g(x)=1$ if and only if $x=1$.

All in all it looks like you're moving between $(\mathbb{R},+)$ and $(\mathbb{R}^+,\cdot)$ with $\log$ and $\exp$. So a nice question.

I'm sure there's more to say.