Áp dụng BĐT Cauchy ta có: xy\le\dfrac{(x+y)^2}{4}=\dfrac{1}{4}.
Ta có:
P=x^2y^2+\dfrac{1}{x^2y^2}+2
=256x^2y^2+\dfrac{1}{x^2y^2}-255x^2y^2+2
\ge2\sqrt{256x^2y^2.\dfrac{1}{x^2y^2}}-255.\dfrac{1}{16}+2
=\dfrac{289}{16}
Dấu bằng xảy ra khi: x=y=\dfrac{1}{2}