15th Junior Turkish Mathematical Olympiad

Let $p=ab$ and $s=a+b$. We want to show that $p^2(s^2-2p-2)\ge s(p-1)$, or equivalently, $p^2{s}^2-(p-1)s-2p^2(p+1)\ge 0$ when $s^2\ge 4p>0$. If $0<p\le 1$, then $p^{2}s-(p-1)\ge p^2\cdot 2\sqrt{p}+(1-p)>0$, and if $p\ge 1$, then $p^{2}s-(p-1)\ge (p^2\cdot 2\sqrt{p}-p)+1>0$. Therefore, $$\begin{array}{rl}{p}^2{s}^2-(p-1)s-2p^2(p+1)& =s(p^{2}s-(p-1))-2p^2(p+1)\\ & \ge 2\sqrt{p}(p^2\cdot 2\sqrt{p}-p+1)-2p^2(p+1)\\ & =2p^3-2p\sqrt{p}-2p^2+2\sqrt{p}\\ & =2\sqrt{p}(p-1)(p\sqrt{p}-1)\\ & \ge 0.\end{array}$$