Relationship between central divisor pairs Define Central Divisors as a pair of natural numbers (a,b), a $<=$ b, that are
closer (in the Euclidean distance sense) to their geometric mean $\sqrt{ab}$ than
any other divisors of the product ab.  For example, (18,20) are central
divisors, but (6,74) are not.  The values of a are OEIS A033676 for n = ab.
I've noticed the following behavior:
Proposition.  If
$$\sqrt{a} > \sqrt{b} - 1,$$
then (a,b) are central divisors.  The inequality is a sufficient, but not
necessary, condition for central divisors, as seen with (12,37).
The inequality is equivalent to
$$a > \left(-1 + \sqrt{1 + 4\sqrt{ab}}\right)^2/4$$
I've tested this numerically without exception, but of course that's not a
proof.  My question is: How can the above proposition be proven?
 A: Here's the correct answer based on the hint by @DanielFischer, including
intermediate steps.  Let a,b,c,d be natural numbers such that ab = cd
and
$$a < c \le d < b;\tag{1}$$
that is, (a,b) are not Central Divisors according to the definition in the
original posting.  Then b > d and c > a, so
$$b + c > d + a$$
$$b - a > d - c$$
$$(b - a)^2 > (d - c)^2\tag{2}$$
Since 4ab = 4cd, we have
$$(b + a)^2 - (b - a)^2 = (d + c)^2 - (d - c)^2\tag{3}$$
From (2) and (3), it follows that
$$(b + a)^2 - (b - a)^2 + (b - a)^2 > (d + c)^2 - (d - c)^2 + (d - c)^2$$
$$(b + a)^2 > (d + c)^2$$
$$b + a > d + c$$
$$d + c \le b + a - 1\tag{4}$$
Assume
$$\sqrt{a} > \sqrt{b} - 1$$
Then
$$\sqrt{b} - \sqrt{a} < 1$$
Squaring both sides, we have
$$b - 2\sqrt{ab} + a < 1$$
$$b + a - 1 < 2\sqrt{ab}$$
By (4),
$$d + c < 2\sqrt{ab} = 2\sqrt{cd}$$
$$d - 2\sqrt{cd} + c < 0$$
$$(\sqrt{d} - \sqrt{c})^2 < 0,$$
a contradiction.  Therefore, there are no solutions for c and d satisfying
(1), and (a,b) are Central Divisors according to the definition in the
original posting.
A: I gave some more thought to the question above and believe I have found the
answer.  From the definition above, (a,b) are central divisors iff ab is not
the product of two factors both of which are smaller than b.  If b is prime,
then the proposition holds trivially for all values of a $\le$ b since
the multiples of b up to $b^2$ cannot be created from two factors both smaller
than b.  For composite b $\le$ 10, the proposition can be seen to hold by
inspection.
Otherwise, write the inequality from the original post as
$$(\sqrt{b} - 1)^2 \lt a \le b$$
Let b $\ge$ 12 be composite and let k be a proper divisor of b.  Then we can
write
$$ab = (ka)(b/k)$$
Assuming both factors on the right are less than b, we have
$$ka \lt b$$ and $$b/k \lt b$$
The latter inequality is always true since k $\ge$ 2, so the assumption is
equivalent to
$$ka \lt b$$ and $$ka \ge 2a$$
We have
$$b \gt (2 + \sqrt{2})^2 = 11.66$$
$$\sqrt{b} \gt 2 + \sqrt{2}$$
This is a solution to the following quadratic inequality:
$$(\sqrt{b})^2 - 4\sqrt{b} + 2 \gt 0$$
$$2b - 4\sqrt{b} + 2 \gt b$$
$$2(\sqrt{b} - 1)^2 \gt b$$
Then
$$ka \ge 2a \gt 2(\sqrt{b} - 1)^2 \gt b,$$
which is a contradiction to the previous assumption that ka $\lt$ b.  Therefore, ab cannot be the
product of two factors both less than b, QED.
A: As an example you give $(12,37)$, but we don't have $\sqrt{12}>\sqrt{37}-1$!
Let us note $n=ab$ where $a\leq b$. So we have $b=n/a$.
It is easy to see that
\begin{equation}
a+b\geq 2\sqrt{n}\tag{1}
\end{equation}
Now let us suppose that
\begin{eqnarray*}
\sqrt{a}>\sqrt{b}-1 
&\Longleftrightarrow &\sqrt{b}-\sqrt{a}<1\\
&\Longleftrightarrow & b-2\sqrt{ab}+a<1\\
&\Longleftrightarrow & b-2\sqrt{n}+a<1\\
&\Longleftrightarrow & b+a<2\sqrt{n}+1 \tag{2}
\end{eqnarray*}
By combining (1) and (2), we conclude that
$$
2\sqrt{n}\leq a+b<2\sqrt{n}+1
$$.
In other terms $a+b=a+n/a$ is the smallest integer greater or equal than $2\sqrt{n}$ which proves that $(a,b)$ are central divisors.
