Ta có :
( a + b + c )^2 = 3ab + 3bc + 3ca thì a = b = c
=> a^2 + b^2 + c^2 +2ab +2bc + 2ac = 3ab + 3bc + 3ca
=>a^2 + b^2 + c^2 - ab - bc - ac = 0
=> 2(a^2 + b^2 + c^2 - ab - bc - ac) = 0
=>2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac = 0
=> a^2 - 2ab + b^2 + b^2 - 2bc +c^2 + c^2 - 2ac + a^2 = 0
=> (a-b)^2 + (b-c)^2 + (c-a)^2 = 0
=> a-b=0
b-c=0
c-a=0
=> a = b
b = c
c = a
Vậy đpcm
