Áp dụng BĐT Cauchy ta có:
$xy\le\dfrac{(x+y)^2}{4}\le\dfrac{1}{4}$
$P=xy+\dfrac{1}{x^2}+\dfrac{1}{y^2}$
$\ge xy+\dfrac{2}{xy}$
$=32xy+\dfrac{2}{xy}-31xy$
$\ge2\sqrt{32xy.\dfrac{2}{xy}}-31.\dfrac{1}{4}=\dfrac{33}{4}$
$\min P=\dfrac{33}{4} \Leftrightarrow x=y=\dfrac{1}{2}$