证明：∵ a²+b¹+c²=ab+bc+ac
∴ 2(a²+b²+c²)=2(ab+bc+ac)
2a²+2b²+2c²=2ab+2bc+2ac
2a²+2b²+2c²-2ab-2bc-2ac=0
∴ (a²-2ab+b²)+(b²-2bc+c²)+(a²-2ac+c²)=0
∴ (a-b)²+(b-c)²+(a-c)²=0
∴ a-b=0 且b-c=0 且a-c=0
∴ a=b 且 b=c 且 a=c
∴ a=b=c