Question

高中数学不等式的证明,谢谢!1/(1+|a|)+1/(1+|b|)≤1+1/(1+|a+b|)

Answer 1

1/(1+|a|)+1/(1+|b|)=(2+|a|+|b|)/[(1+|a|)(1+|b|)]

1+1/(1+|a+b|)>=1+1/(1+|a|+|b|)=(2+|a|+|b|)/(1+|a|+|b|)

所以 1+1/(|a+b|)>={[(1+|a|)(1+|b|)]/(1+|a|+|b|)}*[1/(1+|a|)+1/(1+|b|)]

因为{[(1+|a|)(1+|b|)]/(1+|a|+|b|)}>=1

所以1+1/(|a+b|)>=[1/(1+|a|)+1/(1+|b|)]