loading此时，函数f(x)有3个极值点，且x2?a；
??2lnx1?a?∴当0?a?1时，x1,x3是函数h(x)?2lnx??1的两个零点，?x?2lnx?3??a?1?0x1a?1?0x3消去ａ有
2x1lnx1?x1?2lnx3?x3令g?x??2xlnx?x，g??x??2lnx?1有零点x?数g?x??2xlnx?x在?0,11,且x1??x3所以函ee??1??1?上递减，在,?????上递增，要证明e??e?x1?x3?22?2??2??x3??x1?g?x3??g??x1?因为g?x1??g?x3?即证g?x1??g??x1?也ee?e??e??2??2??1??x1??0构造函数 F?x??g?x1??g??x1?只需要证明它在x??0,?上
e??e??e??就是证g?x1??g??2?2??2x??2?e??0所以F?x在 (0,1]上
单调递减即可，而F??x??2lnx?2ln??x??2，F???x?????2??ee??x??x??e?单调递增, ?F??x??F???1????0 ?e?∴当0?a?1时，x1?x3?
2e.