a+b+c=1
(a+b+c)^2 = 1
a^2 + b^2 + c^2 + 2ab + 2bc + 2ac = 1..........(1)
又因为(a - b)^2 + (b - c)^2 +(a - c)^2 >= 0
a^2 + b^2 + c^2 >= ab + bc + ac ..............(2)
把(2)代入(1)得
3(ab + bc + ac )<= a^2 + b^2 + c^2 + 2ab + 2bc + 2ac = 1
即 3(ab + bc + ac )<= 1
则 ab + bc + ac <= 1/3
(a-b)^2≥0 a^2+b^2≥2ab
(b-c)^2≥0 b^2+c^2≥2bc
(a-c)^2≥0 a^2+c^2≥2ac
因此a^2+b^2+c^2≥ab+bc+ca
(a+b+c)^2=1=a^2+b^2+c^2+2ab+2bc+2ca≥3ab+3bc+3ca
因此3ab+3bc+3ac≤1
ab+bc+ac≤1/3
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