xét hiệu a^2+b^2+c^2-ab-ac-bc=1/2.2(a^2+b^2+c^2-ab-ac-bc)
=1/2(2a^2+2b^2+2c^2-2ab-2ac-2bc)
=1/2[(a^2-2ab+b^2)+(a^2-2ac+c^2)+(b^2-2bc+c^2)]
=1/2.[(a-b)^2+(a-c)^2+(b-c)^2]
vì (a-b)^2+(a-c)^2+(b-c)^2>=0
nên 1/2.[(a-b)^2+(a-c)^2+(b-c)^2]>=0
hay a^2+b^2+c^2-ab-ac-bc >=0<=> a^2+b^2+c^2>=ab+ac+bc

Xét VT -VP = a^2 + b^2 +c^2 -ab -bc -ca
                      = 1/2 ( 2a^2 + 2b^2 + 2c^2 - 2ab -2bc -2ac )
                      =1/2 ( a^2 -2ab - b^2 ) (b^2 - 2bc + c^2 ) ( a^2 -2ac + c^2 )
                      =1/2 {( a - b )^2 ( b - c )^2 ( a - c )^2}
Vì 1/2 > 0
Và {( a - b )^2 ( b - c )^2 ( a - c )^2} >0 
Thì 1/2 {( a - b )^2 ( b - c )^2 ( a - c )^2}  > 0
=> a^2 + b^2 +c^2 > ab + bc +ca 

2(a^2+b^2+c^2)≥2(ab+bc+ca)
(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)≥0
(a-b)^2+(b-c)^2+(c-a)^2≥0