Question

Cho x,y>0 tm: x+y=1

CMR: \(\left(x^2+\dfrac{1}{y^2}\right)\left(y^2+\dfrac{1}{x^2}\right)\ge18\dfrac{1}{16}\)

Answer 1

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