AustriaMO2013

We clear denominators to get $$\begin{array}{rl}& a(a+1)+b(b+1)\le (a+1)(b+1),\\ \iff & a^2+a+b^2+b\le ab+a+b+1,\\ \iff & a^2-a+b^2-b\le ab-a-b+1,\\ \iff & a(a-1)+b(b-1)\le (a-1)(b-1),\\ \iff & (1-a)(1-b)+a(1-a)+b(1-b)\ge 0.\end{array}$$ The three terms on the left-hand side of the last inequality are clearly all positive or zero for $0\le a,b\le 1$. For equality to hold, all three terms have to be zero, that is, $a=1$ or $b=1$ and $a,b\in \{0,1\}$. This gives the three pairs $(a,b)=(1,0)$, $(a,b)=(0,1)$ and $(a,b)=(1,1)$.