22nd Junior Turkish Mathematical Olympiad

Using the given equation, we get $$\begin{array}{rl}0& \ge (ab+1{)}^2+(a+b{)}^2-2(a+b)(a^2-ab+b^2+1)\\ & =a^2{b}^2+4ab+1-2a^3-2b^3+a^2+b^2-2a-2b\\ & =(a^2-2b+1)(b^2-2a+1)\end{array}$$ Since $a>b$, we get $a^2-2b+1>b^2-2b+1=(b-1{)}^2\ge 0$ and $b^2-2a+1\le 0$. Therefore, we have $(b-1{)}^2\le 2(a-b)$ or $$\frac{\sqrt{a-b}}{b-1}\ge \frac{1}{\sqrt{2}}$$ The equality holds at $(a,b)=(5/2,2)$.