cho biểu thức trên = P
\(P=\left(xy\right)^2+\dfrac{1}{\left(xy\right)^2}+2=256\left(xy\right)^2+\dfrac{1}{\left(xy\right)^2}+2-255\left(xy\right)^2< =>P\ge34-255\left(xy\right)^2\)
ta lại có \(x+y\ge2\sqrt{xy}=>1\ge2\sqrt{xy}=>\dfrac{1}{16}\ge\left(xy\right)^2\)
=> \(P\ge34-\dfrac{255}{16}=18\dfrac{1}{16}\)
Dấu = xảy ra khi x=y=1/2