Question

已知a+b+c=1,求证ab+ac+bc<=1/3

RT

Answer 1

证明：因为：a+b+c=1,所以：（a+b+c)^2=1, （a+b+c) ^2=a^2+b^2+c^2+2(ab+bc+ca)=1,再因为：a^2+b^2+c^2>=ab+bc+ca,所以：3(ab+ac+bc)<=1 即：ab+ac+bc<=1/3

Answer 2

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

Answer 3

证明：a+b+c=1,有（a+b+c)^2=1,展开式子有a*a+b*b+c*c+2(ab+bc+ca)=1,又由基本不等式a*a+b*b+c*c>=ab+bc+ca,代入上式即得所求！

Answer 4

(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc=1a^2+b^2>=2aba^2+c^2>=2acb^2+c^2>=2bc2(a^2+b^2+c^2)>=2ab+2ac+2bca^2+b^2+c^2>=ab+ac+bc3(ab+ac+bc)<=1ab+ac+bc<=1/3