(a+b+c)^2=3
a^2+b^2+c^2+2ab+2bc+2ca=3
a^2+b^2+c^2=1
2ab+2bc+2ca=2
所以2a^2+2b^2+2c^2=2ab+2bc+2ca
2a^2+2b^2+2c^2-2ab-2bc-2ca=0
(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
所以(a-b)^2=0,(b-c)^2=0,(c-a)^2=0
a-b=0,b-c=0,c-a=0
所以a=b=c