Ta có: $1\geq x+y\geq2\sqrt{xy}\Rightarrow1\geq4xy\Rightarrow \frac{1}{xy}\geq4$Ta có:
$P\geq2\sqrt{\frac{1}{xy}}\sqrt{1+x^2y^2}=2\sqrt{\frac{1}{xy}+xy}$
Mà $\frac{1}{xy}+xy=\frac{15}{16}.\frac{1}{xy}+\frac{1}{16xy}+xy\geq \frac{15}{16}.4+2\sqrt{\frac{1}{16xy}.xy}=\frac{15}{16}.4+\frac{2}{4}=\frac{17}{4}$
$\Rightarrow P\geq2.\frac{\sqrt{17}}{2}=\sqrt{17}\Leftrightarrow x=y=\frac{1}{2}$.