, 2003

• It is clear that the function $f(x) = x/2$, $x \in \mathbb{R}^+$, is a function belonging to $F$. Thus $\alpha \leq 1/2$.
• Let $f$ be an arbitrary function in $F$. It is easy to see that
$$f(x) \geq x/3 \quad \forall x \in \mathbb{R}^+. \qquad (1)$$
Consider the sequence of numbers $\{\alpha_n\}$ defined by:
$$\alpha_1 = 1/3 \quad \text{and} \quad \alpha_{n+1} = (2\alpha_n^2 + 1)/3 \quad \forall n = 1, 2, 3, \dots$$
By induction on $n$, we shall prove that $\forall n \in \mathbb{N}^+$, we have
$$f(x) \geq \alpha_n x \quad \forall x \in \mathbb{R}^+. \qquad (2)$$
Indeed, (1) shows that we have (2) when $n = 1$.
Suppose that (2) holds for $n = k$. Then:
$$f(x) \geq \alpha_k f(2x/3) + (x/3) \geq \alpha_k \cdot \alpha_k(2x/3) + (x/3) = ((2\alpha_k^2 + 1)/3)x = \alpha_{k+1}x \quad \forall x \in \mathbb{R}^+$$
so (2) holds for $n = k + 1$. So, by induction, (2) is true.
Now we shall prove that $\lim \alpha_n = 1/2$.
Indeed, at first, by induction on $n$, it is easy to prove that the sequence $\{\alpha_n\}$ is bounded above by $1/2$. Therefore,
$$\alpha_{n+1} - \alpha_n = (1/3)(\alpha_n - 1)(2\alpha_n - 1) > 0,$$
it shows that $\{\alpha_n\}$ is an increasing sequence. So $\{\alpha_n\}$ is a convergent sequence.
Passing to the limit, with the remark that $\alpha_n < 1/2$, we find that $\lim \alpha_n = 1/2$, and (2) implies that $f(x) \geq x/2 \quad \forall x \in \mathbb{R}^+$.
* Consequently, the answer to the problem is $\alpha = 1/2$.