Xét: a² + b² + c² - ab - bc - ac
= 1/2.(2a²+2b²+2c²-2ab-2bc-2ac)
= 1/2.[(a²-2ab+b²)+(b²-2bc+c²)+(c²-2ac+a²)]
= 1/2.[(a-b)²+(b-c)²+(c-a)²]
Do (a-b)² ≥ 0
(b-c)² ≥ 0
(a-c)² ≥ 0
=> (a-b)² + (b-c)² + (c-a)² ≥ 0
=> 1/2.[(a-b)² + (b-c)² + (c-a)²] ≥ 0
Vậy a² + b² + c² ≥ ab + bc + ac.