Question

cho x,y>=0 TM x+y=1

tim gtnn của M biết M = (x\(^2\) + \(\dfrac{1}{y^2}\))(y\(^2\)+\(\dfrac{1}{x^2}\))

Answer 1

\(...\Leftrightarrow M=x^2y^2+\dfrac{1}{x^2y^2}+2\left(a\right)\)

Áp dụng Bất đẳng thức Cauchy:

\(1=x+y\ge2\sqrt{xy}\Leftrightarrow xy\le\dfrac{1}{4}\)

\(\left(a\right)\Leftrightarrow M=\left(x^2y^2+\dfrac{1}{256x^2y^2}\right)+\dfrac{255}{256x^2y^2}+2\ge2\sqrt{\dfrac{1}{16^2}+}\dfrac{255}{256.\dfrac{1}{16}}+2\left(vì.xy\le\dfrac{1}{4}\right)\)

\(\Leftrightarrow M\ge\dfrac{17}{8}+\dfrac{255}{16}=\dfrac{289}{16}\)

Vậy \(GTNN\left(M\right)=\dfrac{289}{16}\left(tại.x=y=\dfrac{1}{2}\right)\)