a^3+b^3+c^3-3abc =(a+b)^3+c^3-3a^2b-3ab^2-3abc =(a+b+c)[(a+b)^2-(a+b)c+c^2]-3ab(a+b+c) =(a+b+c)[a^2+b^2+c^2-ab-bc-ca] =1/2*(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2] a、b、c均为正数， 故：a^3+b^3+c^3=3abc a^3+b^3+c^3-3abc=0 (a-b)^2+(b-c)^2+(c-a)^2=0 a=b=c
答：a+b=1, c+d=1 (a+b)(c+d) = (ac+bd)+(ad+bc) = 1 ac+bd > 1 ==> ad +bc a,b,c,d中至少有一...详情>>
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