(a + b + c)^2 = a^2 + ab + ac + ab + b^2 + bc + ac + bc + c^2
= a^2 + b^2 + c^2 + 2ab + 2 ac + 2bc
===> a^2 + b^2 + c^2 = (a + b + c)^2 - (2ab + 2 ac + 2bc)
===> a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + ac + bc)
===> (a^2 + b^2 + c^2) = (a + b + c)^2 - 2(ab + ac + bc)
For all real a, b and c, a^2 + b^2 + c^2 ≥ 0 and (a + b + c)^2 ≥ 0.
Therefore, a^2 + b^2 + c^2 ≥ 2(ab + ac + bc) ===> a^2 + b^2 + c^2 ≥ ab + ac + bc
Đây ạ
