不等式已知a,b∈R,求证：[|a+b|/(1+|a+b|)]≤[|a|/(1+|a|]+[|b|/(1+|b|)].

Question

不等式 已知a,b∈R,求证 ：[|a+b|/(1+|a+b|)]≤[|a|/(1+|a|]+[|b|/(1+|b|)].

Answer 1

构造函数f(x)=x/(1+x),
则f(x)=x/(1+x)=(x+1-1)/(1+x)=1-1/(1+x).
显然x>0时,1/(1+x)递减,-1/(1+x)递增,所以f(x)函数是增函数.
因为|a+b|≤(|a|+|b|)
所以f(|a+b|)≤f(|a|+|b|)
即|a+b|/(1+|a+b|)≤(|a|+|b|)/(1+|a|+|b|)
又因(|a|+|b|)/(1+|a|+|b|)=(|a|)/(1+|a|+|b|)+(|b|)/(1+|a|+|b|)
≤|a|/(1+|a|]+[|b|/(1+|b|),
∴[|a+b|/(1+|a+b|)]≤[|a|/(1+|a|]+[|b|/(1+|b|)].