Question

cho x,y>0; x+y=1. CMR (1+1/x).(1+1/y)>=9

Answer 1

Lời giải:

Áp dụng BĐT Cô-si ta có:

\((1+\frac{1}{x})(1+\frac{1}{y})=\frac{x+1}{x}.\frac{y+1}{y}=\frac{(x+1)(y+1)}{xy}\)

\(=\frac{(x+x+y)(y+x+y)}{xy}\geq \frac{3.\sqrt[3]{x^2y}.3\sqrt[3]{xy^2}}{xy}=\frac{9xy}{xy}=9\) 

Vậy ta có đpcm

Dấu "=" xảy ra khi $x=y=\frac{1}{2}

Answer 2

Cách 2:

\((1+\frac{1}{x})(1+\frac{1}{y})=1+\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}=1+\frac{x+y}{xy}+\frac{1}{xy}\)

\(=1+\frac{2}{xy}\)

Áp dụng BĐT Cô-si:

$xy\leq \frac{(x+y)^2}{4}=\frac{1}{4}$

$\Rightarrow \frac{2}{xy}\geq 8$

$\Rightarrow (1+\frac{1}{x})(1+\frac{1}{y})\geq 1+8=9$

Ta có đpcm

Dấu "=" xảy ra khi $x=y=\frac{1}{2}$