Question

函数y=

x2+2

x2+1

的最小值为______．

Answer 1

∵y=

x2+2

x2+1

=

x2+1+1

x2+1

=

x2+1

+

1

x2+1

≥2

x2+1 × 1 x2+1

=2，当且仅当

x2+1

=

1

x2+1

，即x=0时取等号．
∴y的最小值为2．
故答案为2．