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證明:設f(x)=xx+m(x>0),則f′(x)=m(x+m)2>0∴f(x)在(0,+∞)上為增函數.在△ABC中,a+b>c,則a+ba+b+m>cc+m.∴cc+m<aa+b+m+ba+b+m<aa+m+bb+m.∴原不等式成立.
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