Question

cho x,y>0 và xy=1 cùng\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{2}{x+y}\ge3\)

Answer 1

Ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{2}{x+y}=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x+y}+\dfrac{1}{x+y}\)

Áp dung BĐT Cô - si cho 2 số không âm, ta có:

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

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

Cộng vế theo vế BĐT ta được:

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x+y}+\dfrac{1}{x+y}\ge2+1=3\)

\(\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{2}{x+y}\ge3\)

Dấu " =" xảy ra khi: x = y = 1