Romanian Mathematical Olympiad

a) $xy-x-y+1=(x-1)(y-1)$.

b) Both members of the inequality are positive, so we can square and get the equivalent form $a^2+b^2+2ab>1+2ab+a^2{b}^2$, that is $(a^2-1)(b^2-1)<0$. This shows that $a^2-1<0$ or $b^2-1<0$. Since $a$ and $b$ are integers, this yields $a=0$ or $b=0$.