67th Romanian Mathematical Olympiad

a) It is clear that for any real numbers $u,v$, we have $|u+v|+|u-v|\in \{\pm 2u,\pm 2v\}$, and thus $|\alpha x+\beta y|+|\alpha x-\beta y|\le \max \{2|\alpha |,2|\beta |\}$. As we get the equality for $x=y=1$, the quantity on the right hand side is the maximal value.

b) It is not difficult to check the equality $|\alpha x+\beta y{|}^2+|\alpha x-\beta y{|}^2=2|\alpha x{|}^2+2|\beta y{|}^2$. On the other hand, we have $$(|\alpha x+\beta y|+|\alpha x-\beta y|{)}^2\le 2\left(|\alpha x+\beta y{|}^2+|\alpha x-\beta y{|}^2\right),$$ and thus $|\alpha x+\beta y|+|\alpha x-\beta y|\le 2\sqrt{{\alpha }^2+{\beta }^2}$. We get the maximal value $2\sqrt{{\alpha }^2+{\beta }^2}$ for $x=1$ and $y=i$.