có a+b+c= 1
=> (a+b+c)²= 1
<=> a²+ b²+ c²+ 2ab+ 2ac+ 2bc= 1
Có (a-b)²+ (b-c)²+ (c-a)² ≥0
<=> a² -2ab+ b²+ b²-2bc+ c²+ c²- 2ac+ a² ≥0
<=> 2 (a²+ b²+ c²- ab- bc- ac)≥ 0
<=> a²+ b²+ c²- ab- bc- ac≥ 0
<=> a²+ b²+ c²≥ ab+ bc+ ac
<=> a²+ b²+ c²+ 1≥ ab+ bc+ ac+ a²+ b²+ c²+ 2ab+ 2ac+ 2bc
(vì a²+ b²+ c²+ 2ab+ 2ac+ 2bc= 1)
<=> a²+ b²+ c²+ 1 - a²- b²- c²≥ 3ab+ 3ac+ 3bc
<=> 1 ≥ 3ab+ 3ac+ 3bc
<=> 1/3 ≥ ab+ bc+ ac
Mik chỉ cm đc ab+ bc+ ac ≤ 1/3 thôi!!!