已知a+b+c=1,求证:ab+bc+ac<=1⼀3

2024-11-27 21:08:19
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回答(1):

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