Ta có: \(1\ge x+y\ge2\sqrt{xy}\Rightarrow1\ge4xy\Rightarrow\frac{1}{xy}\ge4\)
\(\Rightarrow P\ge2\sqrt{\frac{1}{xy}}\cdot\sqrt{1+x^2y^2}=2\sqrt{\frac{1}{xy}+xy}\)
Mà \(\frac{1}{xy}+xy=\frac{15}{16}\cdot\frac{1}{xy}+\frac{1}{16xy}+xy\)
\(\ge\frac{15}{16}\cdot4+2\sqrt{\frac{1}{16xy}\cdot xy}=\frac{15}{16}\cdot4+\frac{2}{4}=\frac{17}{4}\)
\(\Rightarrow P\ge2\cdot\frac{\sqrt{17}}{2}=\sqrt{17}\) xảy ra khi \(x=y=\frac{1}{2}\)