Question

设a＞0，b＞0且ab－(a+b)≥1，则

A.a+b≥2(+1)                                            B.a+b≤+1

C.a－b≤(+1)²                                           D.a+b＞2(+1)

Answer 1

解析：∵≤，∴ab≤(a+b)².

∴(a+b)²－(a+b)≥ab－(a+b)≥1.

∴(a+b)²－4(a+b)－4≥0.

∴a+b≤2－2或a+b≥2+2.

∵a＞0，b＞0，∴a+b≥2+2成立.

答案：A