cho a + b+c > 0 thỏa mãn a + b +c = 1 . Chứng minh rằng ab + bc + ca >= 1/3

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!!!