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?2lnx1???即有??2lnx3???a?1?0x1,消去a有2x1lnx1?x1?2x3lnx3?x3
a?1?0x31e,且x1?令g(x)?2xlnx?x,g'(x)?2lnx?1有零点x?1e?x3
∴函数g(x)?2xlnx?x在(0,1e)上递减,在(2e2e1e,??)上递增
要证明 x1?x3?2e?x3??x1?g(x3)?g(2e?x1)
?g?x1??g?x3? ?即证g(x1)?g(构造函数F?x??g(x)?g(?x1)?g(x1)?g(2e?x1)?0
2?1??x),?F?=0 ???e?e?2e?x)?2,
只需要证明x?(0,1e]单调递减即可.而F??x??2lnx?2ln(2(F''?x??x(2e2e?2x)?0 ?F??x?在(0,?x)2e1?1?]上单调递增, ?F??x??F?????0 e?e?∴当0?a?1时,x1?x3?
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