$S=\dfrac{1-y}{\sqrt y}+\dfrac{1-x}{\sqrt x}= (\dfrac{1}{\sqrt x}+\dfrac{1}{\sqrt y})-(\sqrt x + \sqrt y)$
Mặt khác $S=\dfrac{x}{\sqrt y}+\dfrac{y}{\sqrt x}=(\dfrac{x}{\sqrt y} +\sqrt y) + (\dfrac{y}{\sqrt x} +\sqrt x) -(\sqrt x + \sqrt y)$
$\ge 2\sqrt x +2\sqrt y -(\sqrt x + \sqrt y) = \sqrt x + \sqrt y$
Từ 2 điều đó $\Rightarrow2S \ge \dfrac{1}{\sqrt x}+\dfrac{1}{\sqrt y}\ge \dfrac{2}{\sqrt[4]{xy}} \ge \dfrac{2}{\sqrt{\dfrac{x+y}{2}}}=2\sqrt 2$
$\Rightarrow S\ge \sqrt 2 \Rightarrow \min S=\sqrt 2 \Leftrightarrow (x;\ y) =(\dfrac{1}{2};\ \dfrac{1}{2})$