??4sin?
所以曲线C的极坐标方程为??4sin?
（II）将l的参数方程??x?2?tcos?222代入x?y?4y?0，得t?4(sin?cos)t?4?0
?y?4?tsin???16(sin??cos?)2?16?16sin2??0?t1?t2??4(sin??cos?)∴? 所以sin2??0，又0????，
?t1t2?4?所以??(0,?2)，且t1?0,t2?0
所以|MA|?|MB|?|t1|?|t2|?|t1?t2|?4(sin??cos?)?42sin(???4)
由??(0,?2)，得????3?2??(,)，所以?sin(??)?1. 44424故|MA|?|MB|的取值范围是(4,42]
23. 证明（Ⅰ）∵a2?b2?2ab,b2?c2?2bc,c2?a2?2ca，三式相加可得
a2?b2?c2?ab?bc?ca
∴(a?b?c)?a?b?c?2ab?2bc?2ca?(ab?bc?ca)?2(ab?bc?ca)
2222?3(ab?bc?ca)?9
又a、b、c均为正整数，∴a?b?c?3成立． （Ⅱ）：a、b?R，a?b?1，∴a?2ab?b?1，
?2211a2?2ab?b2a2?2ab?b2?1)(?1) ∴(2?1)(2?1)?(aba2b22bb22aa22a2b2a2b?(?2)(?2)=5+??5?2??9
aabbbaba当且仅当
2a2b1?，即a?b?时，“=”成立 ba2