Question

Cho \(x,y,z>0\). C/m:
a, \(\left(x+y\right)\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\ge0\)

b, \(\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right)\ge64\) với \(x+y+z=1\)

Answer 1

a )

Sử dụng Cô-si , ta có :

\(x+y\ge2\sqrt{xy}\) (1)

\(\dfrac{1}{x}+\dfrac{1}{y}\ge2\sqrt{\dfrac{1}{x}.\dfrac{1}{y}}\) (2)

Nhân cả vế (1) vế (2) lại ta có :

\(\left(x+y\right)\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\ge2\sqrt{xy}.2\sqrt{\dfrac{1}{x}.\dfrac{1}{y}}=4\)

\(\LeftrightarrowĐPCM.\)

Answer 2

Câu b trên mạng đầy :v

Answer 3

\(\left(x+y\right)\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=1+\dfrac{x}{y}+\dfrac{y}{x}+1\ge1+2+1=4\ge0\)