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