( a+ b+c) ^2 >= 3(ab+ bc+ ca)
=> a^2 + b^2 + c^2 + 2ab+ 2bc + 2ac >= 3ab + 3bc + 3ac
=> a^2 + b^2 + c^2 + 2ab + 2bc+ 2ac - 3ab - 3bc - 3ac >=0
=> a^2 + b^2 + c^2 - ab- bc - ac >=0
=> 2( a^2 + b^2 + c^2 - ab-bc-ac) >=0
=> 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac >=0
=> a^2 - 2ab + b^2 + b^2 -2bc + c^2 + c^2 - 2ac + a^2 >=0
=> (a-b)^2 +(b-c)^2 +(c-a)^2 >=0
vì (a-b)^2 >=0
(b-c)^2 >=0
(c-a)^2 >=0
nên (a-b)^2 +(b-c)^2 +(c-a)^2 >=0
hay (a+b+c)^2 >= 3(ab+bc+ca)
Xét (a+b+c)2=a2+b2+c2+2ab+2ac+2bc
Xét 3(ab+bc+ca) = 3ab + 3bc + 3ca = 2ab+2bc+2ac + ab+bc+ac
Vậy (a+b+c)2\(\ge\)3(ab+ac+bc)
Dấu "=" xảy ra khi a=b=c
