Minimum $a+b$ such that $ab=M$ This question is related to the fleablood's answer to another question.
Let $ab=M$ where $0<a,b,M\in\mathbb N$. He wanted to show that the sum $a+b$ is the smallest when $a,b$ are closest together. But he defined $a=\sqrt{M}r$, $b=\sqrt{M}/r$ with $r=1+e$ (where I suppose that $0<e\in\mathbb R$).
My question. Is it always possible to define the positive integers in $ab=M$ as $a=\sqrt{M}r$ and $b=\sqrt{M}/r$ with $r=1+e$? How would you show it?
Thanks in advance.

Since $ab = M$, by setting $a = \sqrt{M}r$ we immediately get $b = \frac{\sqrt{M}}{r}$, for some $r > 0$.
Without loss of generality, we then assume $a \geq b$ (in other words, because everything in the problem is symmetrical we'll get the same solution if $a \geq b$ or $b \geq a$, so we pick one of them arbitrarily). The implication of that is:
$\sqrt{M}r \geq \frac{\sqrt{M}}{r} \implies r^2 \geq 1 \implies r \geq 1$
Then, since $r \geq 1$, if we set $r = 1 + e$ then that immediately implies $e \geq 0$.
It's important that $e = 0$ is a possibility, since for $M = 1$ the only factorisation is $a = b = 1$ which gives $r = 1$ and $e = 0$ as a result, and fleablood's solution does cover this.
