问题
选择题
已知x,y∈R+,且xy=1,则(1+
|
答案
(1+
)(1+1 x
)=1+1 y
+1 x
+1 y
=2+1 xy
+1 x
=2+1 y
=2+(x+y)x+y xy
因为x,y∈R+,且xy=1,所以2+(x+y)≥2+2
=2+2=4,当且仅当x=y=1时取等号.xy
所以(1+
)(1+1 x
)的最小值为4,1 y
故选A.
已知x,y∈R+,且xy=1,则(1+
|
(1+
)(1+1 x
)=1+1 y
+1 x
+1 y
=2+1 xy
+1 x
=2+1 y
=2+(x+y)x+y xy
因为x,y∈R+,且xy=1,所以2+(x+y)≥2+2
=2+2=4,当且仅当x=y=1时取等号.xy
所以(1+
)(1+1 x
)的最小值为4,1 y
故选A.