26th Hellenic Mathematical Olympiad

(α) To each fraction of $A$ of the form $\frac{2\nu -1}{2\nu +2}$, $\nu =1,2,\dots ,299$, corresponds a fraction from $B$ of the form $\frac{2\nu }{2\nu +3}$, $\nu =1,2,\dots ,299$. Since $$0<\frac{2\nu -1}{2\nu +2}<\frac{2\nu }{2\nu +3}\text{for every }\nu \in {\mathbb{N}}^{\ast },$$ and so for $\nu =1,2,\dots ,299$, by multiplying by parts the above 299 inequalities we obtain $A<B$.

(β) Since $A>0$, from $A<B$ we get: $$A^2<A\cdot B=\frac{1\cdot 2\cdot 3}{599\cdot 600\cdot 601}<\frac{1}{100\cdot 599^2}=\frac{1}{5990^2}\Rightarrow A<\frac{1}{5990}.$$