đặt A=\(\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)=\left(1+\dfrac{x+y}{x}\right)\left(1+\dfrac{x+y}{y}\right)\)
=\(\left(1+\dfrac{x}{x}+\dfrac{y}{x}\right)\left(1+\dfrac{y}{y}+\dfrac{x}{y}\right)\)
=\(\left(2+\dfrac{y}{x}\right)\left(2+\dfrac{x}{y}\right)\)
=\(4+\dfrac{2x}{y}+\dfrac{2y}{x}+\dfrac{xy}{yx}\)
=\(4+2\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+1\)
xét
Áp dụng BĐT cô si cho 2 số ko âm ta có
\(\dfrac{x}{y}+\dfrac{y}{x}\ge2\sqrt{\dfrac{x}{y}.\dfrac{y}{x}}=2\)
=> A ≥ 4+4+1
=> A ≥ 9 (đpcm)
Ta có :
\(A=\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)=\left(1+\dfrac{x+y}{x}\right)\left(1+\dfrac{x+y}{y}\right)\)
\(A=\left(2+\dfrac{y}{x}\right)\left(2+\dfrac{x}{y}\right)\)
\(A=4+\dfrac{2x}{y}+\dfrac{2y}{x}+1=5+2\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)
Theo BĐT cô si \(\dfrac{x}{y}+\dfrac{y}{x}\ge2\sqrt{\dfrac{x}{y}\times\dfrac{y}{x}}=2\)
Do đó : \(A\ge5+2.2=9\)
Dấu \("="\) xảy ra khi \(x=y=\dfrac{1}{2}\)
