Find all binary operations $\diamond: \mathbb{R}_{>0} \times \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ (meaning $\diamond$ takes pairs of positive real numbers to positive real numbers) such that for any real numbers $a, b, c>0$,
- the equation $a \diamond(b \diamond c)=(a \diamond b) \cdot c$ holds; and
- if $a \geq 1$ then $a \diamond a \geq 1$.
The answer is only multiplication and division, which both obviously work.
We present two approaches, one appealing to theorems on Cauchy's functional equation, and one which avoids it.
₪ First solution using Cauchy FE. We prove:
Claim - We have $a \diamond b=a f(b)$ where $f$ is some involutive and totally multiplicative function. (In fact, this classifies all functions satisfying the first condition completely.)

Proof. Let $P(a, b, c)$ denote the assertion $a \diamond(b \diamond c)=(a \diamond b) \cdot c$.
- Note that for any $x$, the function $y \mapsto x \diamond y$ is injective, because if $x \diamond y_{1}=x \diamond y_{2}$ then take $P\left(1, x, y_{i}\right)$ to get $y_{1}=y_{2}$.
- Take $P(1, x, 1)$ and injectivity to get $x \diamond 1=x$.
- Take $P(1,1, y)$ to get $1 \diamond(1 \diamond y)=y$.
- Take $P(x, 1,1 \diamond y)$ to get
$$
x \diamond y=x \cdot(1 \diamond y) .
$$

Henceforth let us define $f(y)=1 \diamond y$, so $f(1)=1, f$ is involutive and
$$
x \diamond y=x f(y) \text {. }
$$

Plugging this into the original condition now gives $f(b f(c))=f(b) c$, which (since $f$ is an involution) gives $f$ completely multiplicative.

In particular, $f(1)=1$. We are now interested only in the second condition, which reads $f(x) \geq 1 / x$ for $x \geq 1$.
Define the function
$$
g(t)=\log f\left(e^{t}\right)
$$
so that $g$ is additive, and also $g(t) \geq-t$ for all $t \geq 0$. We appeal to the following theorem:
Lemma
If $h: \mathbb{R} \rightarrow \mathbb{R}$ is an additive function which is not linear, then it is dense in the plane: for any point $\left(x_{0}, y_{0}\right)$ and $\varepsilon>0$ there exists $(x, y)$ such that $h(x)=y$ and $\sqrt{\left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2}}0$, we have $f(a) \in\{1 / a, a\}$.

Proof. WLOG $b>1$, and suppose $f(b)=a \geq 1 / b$ hence $f(a)=b$.
Assume that $a b>1$; we show $a=b$. Note that for integers $m$ and $n$ with $a^{n} b^{m} \geq 1$, we must have
$$
a^{m} b^{n}=f(b)^{m} f(a)^{n}=f\left(a^{n} b^{m}\right) \geq \frac{1}{a^{n} b^{m}} \Longrightarrow(a b)^{m+n} \geq 1
$$
and thus we have arrived at the proposition
$$
m+n<0 \Longrightarrow n \log _{b} a+m<0
$$
for all integers $m$ and $n$. Due to the density of $\mathbb{Q}$ in the real numbers, this can only happen if $\log _{b} a=1$ or $a=b$.
Claim - The function $f$ is continuous.

Proof. Indeed, it's equivalent to show $g(t)=\log f\left(e^{t}\right)$ is continuous, and we have that
$$
|g(t)-g(s)|=\left|\log f\left(e^{t-s}\right)\right|=|t-s|
$$
since $f\left(e^{t-s}\right)=e^{ \pm|t-s|}$. Therefore $g$ is Lipschitz. Hence $g$ continuous, and $f$ is too.
Finally, we have from $f$ multiplicative that
$$
f\left(2^{q}\right)=f(2)^{q}
$$
for every rational number $q$, say. As $f$ is continuous this implies $f(x) \equiv x$ or $f(x) \equiv 1 / x$ identically (depending on whether $f(2)=2$ or $f(2)=1 / 2$, respectively).
Therefore, $a \diamond b=a b$ or $a \diamond b=a \div b$, as needed.
Remark. The Lipschitz condition is one of several other ways to proceed. The point is that if $f(2)=2$ (say), and $x / 2^{q}$ is close to 1 , then $f(x) / 2^{q}=f\left(x / 2^{q}\right)$ is close to 1 , which is enough to force $f(x)=x$ rather than $f(x)=1 / x$.