Let $\mathcal{F}$ be the set of all functions $f : (0,\infty)\to (0,\infty)$ such that $f(3x) \geq f( f(2x) )+x$ for all $x$. Find the largest $A$ such that $f(x) \geq A x$ for all $f\in\mathcal{F}$ and all $x$.
1. Define the set $\mathcal{F}$ as the set of all functions $f : (0,\infty) \to (0,\infty)$ such that $f(3x) \geq f(f(2x)) + x$ for all $x > 0$.
2. We aim to find the largest $A$ such that $f(x) \geq A x$ for all $f \in \mathcal{F}$ and all $x > 0$.
3. Define $\mathcal{G}$ as the set of all $\alpha \in (0,\infty)$ such that $f(x) \geq \alpha x$ for all $f \in \mathcal{F}$ and all $x > 0$.
4. To show that $\mathcal{G}$ is not empty, consider the inequality $f(x) \geq f(f(2x/3)) + x/3$. Since $f(f(2x/3)) > 0$, it follows that $f(x) > x/3$. Hence, $1/3 \in \mathcal{G}$.
5. Consider the function $f(x) = x/2$. This function is in $\mathcal{F}$ because:
   \[
   f(3x) = \frac{3x}{2}, \quad f(f(2x)) = f\left(\frac{2x}{2}\right) = f(x) = \frac{x}{2}
   \]
   Thus,
   \[
   f(3x) = \frac{3x}{2} \geq \frac{x}{2} + x = \frac{3x}{2}
   \]
   This shows that $f(x) = x/2$ satisfies the condition, so $1/2$ is an upper bound for $\mathcal{G}$.
6. Therefore, $A = \sup \mathcal{G} \leq 1/2$ exists and $A \in \mathcal{G}$.
7. For all $f \in \mathcal{F}$ and all $x > 0$, we have:
   \[
   f(x) \geq f(f(2x/3)) + x/3 \geq A f(2x/3) + x/3
   \]
8. Since $f(2x/3) \geq A (2x/3)$, we get:
   \[
   f(x) \geq A \left(\frac{2x}{3}\right) + \frac{x}{3} = \frac{2A x}{3} + \frac{x}{3} = \left(\frac{2A + 1}{3}\right) x
   \]
9. Therefore, $\frac{2A + 1}{3} \in \mathcal{G}$, and hence $\frac{2A + 1}{3} \leq A$.
10. Solving the inequality $\frac{2A + 1}{3} \leq A$:
    \[
    2A + 1 \leq 3A \implies 1 \leq A \implies A \geq 1
    \]
11. Combining this with $A \leq 1/2$, we get $A = 1/2$.

The final answer is $\boxed{\frac{1}{2}}$