Least value of $x+y+z$ where $ax=by=cz$

Question

The following question is a generalization of the case $a=3$, $b=4$, $c=5$ from a MindYourDecisions YouTube video (which I am not going to actually link here).

Given positive integers $a$, $b$, and $c$, what is the smallest possible value of $x+y+z$ where $x$, $y$, and $z$ are positive integers with $ax=by=cz$?

If $a$, $b$, and $c$ are pairwise coprime, then the answer is just $bc+ac+ab$, because $lcm(a,b,c)=abc$.

The case $a=3$, $b=4$, $c=7$ was asked 6 years ago in Least Value Of $x+y+z$.

Answer 1

Start with:

$$x_0 = bc$$ $$y_0 = ac$$ $$z_0 = ab$$

and then, if $x_0$,$y_0$ and $z_0$ share a common prime factor $p$, divide each of them by $p$ and check again. You will end up dividing by the gcd(x_0,y_0,z_0). Thus, it's simply:

$$\min(x+y+z) = \frac{bc+ac+ab}{\gcd(bc,ac,ac)}$$

Answer 2

Let $N = ax = by = cz$. Since $N$ is a multiple of $a, b, c$ it must be a multiple of $\mathrm{lcm}(a, b, c)$. Thus, we should take $$ \begin{align*} x &= N/a \\ y &= N/b \\ z &= N/c \end{align*} $$ The sum will be minimal when $N = \mathrm{lcm}(a, b, c)$.