2003 Vietnamese Mathematical Olympiad

• It is clear that the function $f(x)=x/2$, $x\in {\mathbb{R}}^{+}$, is a function belonging to $F$. Thus $\alpha \le 1/2$. • Let $f$ be an arbitrary function in $F$. It is easy to see that $$f(x)\ge x/3\forall x\in {\mathbb{R}}^{+}.(1)$$ Consider the sequence of numbers $\{{\alpha }_n\}$ defined by: $${\alpha }_1=1/3\text{and}{\alpha }_{n+1}=(2{\alpha }_n^2+1)/3\forall n=1,2,3,\dots$$ By induction on $n$, we shall prove that $\forall n\in {\mathbb{N}}^{+}$, we have $$f(x)\ge {\alpha }_{n}x\forall x\in {\mathbb{R}}^{+}.(2)$$ Indeed, (1) shows that we have (2) when $n=1$. Suppose that (2) holds for $n=k$. Then: $$f(x)\ge {\alpha }_{k}f(2x/3)+(x/3)\ge {\alpha }_k\cdot {\alpha }_k(2x/3)+(x/3)=((2{\alpha }_k^2+1)/3)x={\alpha }_{k+1}x\forall x\in {\mathbb{R}}^{+}$$ 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 $f(x)\ge x/2\forall x\in {\mathbb{R}}^{+}$. * Consequently, the answer to the problem is $\alpha =1/2$.