Question

cho x,y>0 và x+y=2 tìm gtnn của  (2x+1/x)²+(2y+1/y)²+2001 
mik cảm ơn ạ

 

Answer 1

-Áp dụng BĐT AM-GM ta có:

\(xy\le\dfrac{\left(x+y\right)^2}{4}\Leftrightarrow xy\le\dfrac{2^2}{4}=1\)

\(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}=\dfrac{2^2}{2}=2\)

\(A=\left(2x+\dfrac{1}{x}\right)^2+\left(2y+\dfrac{1}{y}\right)^2+2001=4x^2+4+\dfrac{1}{x^2}+4y^2+4+\dfrac{1}{y^2}+2001=4\left(x^2+y^2\right)+\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+2009\ge4.2+2.\dfrac{1}{xy}+2009\ge8+2.\dfrac{1}{1}+2009=2019\)

\(A=2019\Leftrightarrow x=y=1\)

-Vậy \(A_{min}=2019\)