(Vietnam 2003)(M) Let $F$ be the set of functions from $\mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that $f(3 x) \geqslant f(f(2 x)) + x$. Maximize $\alpha$ such that $\forall x \geqslant 0, \forall f \in F, f(x) \geqslant \alpha x$.
$f(x) \geqslant \frac{x}{3}$ so our sup is well defined in $\overline{\mathbb{R}} \cdot \frac{x}{2} \in F$ so $\alpha \leqslant \frac{1}{2}$ now we write

$$
f(3 x) \geqslant f(f(2 x))+x \geqslant\left(2 \alpha^{2}+1\right) x
$$

and by maximality $\alpha \geqslant 2 \alpha^{2}+1$ and thus $\alpha \in\left[\frac{1}{2}, 1\right]$ and so $\alpha=\frac{1}{2}$.