48th Austrian Mathematical Olympiad

$$\frac{a^3+b^3}{a^2+b^2}=(a+b)\frac{a^2-ab+b^2}{a^2+b^2}=1-\frac{ab}{a^2+b^2}.$$ From this the right inequality is evident with equality for $ab=0$, i.e. for $a=0$, $b=1$ and for $a=1$, $b=0$.

The left inequality is equivalent to $$\frac{1}{2}\le 1-\frac{ab}{a^2+b^2}⟺\frac{ab}{a^2+b^2}\le \frac{1}{2}⟺2ab\le a^2+b^2⟺0\le (a-b{)}^2.$$ This inequality is obvious with equality for $a=\frac{1}{2}$. In this case also $b=\frac{1}{2}$.