Question

Cho x,y là 2 số dương: x + y = 1.

Tìm GTNN: M= ( x + \(\dfrac{1}{x}\))\(^2\) + (y +\(\dfrac{1}{y}\))\(^2\).

Giúp mình nha, gấp lắm ^^

Answer 1

\(M=x^2+\dfrac{1}{x^2}+2+y^2+\dfrac{1}{y^2}+2=x^2+y^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}+4\)

ta có \(x+y=1\Rightarrow x^2+y^2=1-2xy\) theo BĐT Cô si: \(xy\le\dfrac{\left(x+y\right)^2}{4}\Rightarrow2xy\le\dfrac{\left(x+y\right)^2}{2}\Rightarrow1-2xy\ge1-\dfrac{1}{2}=\dfrac{1}{2}\)

\(\Rightarrow x^2+y^2\ge\dfrac{1}{2}\)

Áp dụng tiếp BĐT Cô Si :\(\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{2}{xy}\ge\dfrac{2}{\dfrac{\left(x+y\right)^2}{4}}=\dfrac{2}{\dfrac{1}{4}}=8\)

\(\Rightarrow x^2+y^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}+4\ge\dfrac{1}{2}+8+4=\dfrac{25}{2}\)

dấu = xảy ra tại \(x=y=\dfrac{1}{2}\)