Find all binary operations $\diamondsuit: \mathbb R_{>0}\times \mathbb R_{>0}\to \mathbb R_{>0}$ (meaning $\diamondsuit$ takes pairs of positive real numbers to positive real numbers) such that for any real numbers $a, b, c > 0$, 
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[*] the equation $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ holds; and 
[*] if $a\ge 1$ then $a\,\diamondsuit\, a\ge 1$. 
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[i]Evan Chen[/i]
1. **Define the operation and notation:**
   Let $a \diamondsuit b = f(a, b)$. We need to find all functions $f: \mathbb{R}_{>0} \times \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ such that:
   \[
   f(a, f(b, c)) = f(f(a, b), c) \quad \text{for all } a, b, c > 0
   \]
   and
   \[
   a \ge 1 \implies f(a, a) \ge 1.
   \]

2. **Injectivity of $f(a, b)$ with respect to $b$:**
   Suppose $f(a, b_1) = f(a, b_2)$. Then, using the first condition:
   \[
   f(c, f(a, b_1)) = f(c, f(a, b_2)) \implies b_1 f(c, a) = b_2 f(c, a).
   \]
   Since $f(c, a) \neq 0$ for all $c, a > 0$, we must have $b_1 = b_2$. Thus, $f(a, b)$ is injective with respect to $b$.

3. **Define a function $g$:**
   Define $g: \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ by $g(a) = f(1, a)$.

4. **Show that $f(a, 1) = a$:**
   Using the first condition with $b = a$ and $c = 1$:
   \[
   f(b, f(a, 1)) = f(b, a).
   \]
   By injectivity, $f(a, 1) = a$.

5. **Express $f(a, b)$ in terms of $g$:**
   Using the first condition with $a, b, \frac{a}{f(a, b)}$:
   \[
   f\left(a, f\left(b, \frac{a}{f(a, b)}\right)\right) = a \implies f\left(b, \frac{a}{f(a, b)}\right) = 1.
   \]
   Similarly, using $1, b, \frac{1}{f(1, b)}$:
   \[
   f\left(1, f\left(b, \frac{1}{f(1, b)}\right)\right) = 1 \implies f\left(b, \frac{1}{f(1, b)}\right) = 1.
   \]
   Therefore, $\frac{a}{f(a, b)} = \frac{1}{f(1, b)}$, implying $f(a, b) = a f(1, b) = a g(b)$.

6. **Properties of $g$:**
   Using $P(1, 1, c)$:
   \[
   g(g(c)) = c.
   \]
   Using $P(1, b, g(c))$:
   \[
   g(bc) = g(b)g(c).
   \]

7. **Case analysis for $g$:**
   - **Case 1: There exists $a > 1$ such that $g(a) < 1$.**
     By the given inequality, $g(a) \ge \frac{1}{a}$. Also:
     \[
     a = g(g(a)) = \frac{1}{g\left(\frac{1}{g(a)}\right)} \le \frac{1}{g(a)} \implies g(a) \le \frac{1}{a}.
     \]
     Thus, $g(a) = \frac{1}{a}$. For any $x \in \mathbb{R}_{>0}$, let $k$ be such that $a^k \le x < a^{k+1}$. Then:
     \[
     g(x) = \frac{1}{a^k} g\left(\frac{x}{a^k}\right) \ge \frac{1}{x},
     \]
     and
     \[
     g(x) = \frac{\frac{1}{a^{k+1}}}{g\left(\frac{a^{k+1}}{x}\right)} \le \frac{1}{x}.
     \]
     Hence, $g(x) = \frac{1}{x}$ for all $x \in \mathbb{R}_{>0}$.

   - **Case 2: For all $a > 1$, $g(a) > 1$.**
     Since $g$ is multiplicative and strictly increasing, and $g$ is an involution, it must be the identity function.

8. **Conclusion:**
   The solutions are $g(x) = x$ and $g(x) = \frac{1}{x}$. Therefore, the binary operations are:
   \[
   a \diamondsuit b = ab \quad \text{and} \quad a \diamondsuit b = \frac{a}{b}.
   \]

The final answer is $\boxed{a \diamondsuit b = ab}$ and $\boxed{a \diamondsuit b = \frac{a}{b}}$.