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. 【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.)
- 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 
$$ x \diamond y=x f(y) $$
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 $\sqrt{\left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2}}<\varepsilon$.
Applying this lemma with the fact that $g(t) \geq-t$ implies readily that $g$ is linear. In other words, $f$ is of the form $f(x)=x^{r}$ for some fixed real number $r$. It is easy to check $r= \pm 1$ which finishes.