Question

已知|x|≤1,|y|≤1,证明|x+y|≤|1+xy|

Answer 1

证明：

可以利用作差比较法

∵ |1+xy|²-|x+y|²

= (1+xy)²-(x+y)²

=1+2xy+x²y²-(x²+2xy+y²)

=1+x²y²-x²-y²

=1-x²+y²(x²-1)

=(1-x²)(1-y²)

∵ |x|≤1,|y|≤1

∴ 1-x²≥0,1-y²≥0

∴ (1-x²)(1-y²)≥0

∴ |1+xy|²-|x+y|²≥0

∴ |1+xy|²≥|x+y|²

∴ |x+y|≤|1+xy|