∵(√a+ √b+√c)^2=3(√ab+√bc+√ca)
(√a)^2+(√b)^2+(√c)^2+2√ab+2√bc+2√ca-3√ab-3√bc-3√ca=0
(√a)^2+(√b)^2+(√c)^2-√ab-√bc-√ca=0
各項(xiàng)同乘以2,整理后得：
(√a-√b)^2+(√b-√c)^2+(√c-√a)^2=0
∵左邊各項(xiàng)均為(+),各項(xiàng)正數(shù)之和為零,則每一項(xiàng)均為零,即：
(√a-√b)^2=0,(√b-√c)=0,(√c-√a)^2=0.
√a-√b=0,a=b；
√b-√c=0,b=c；
√c-√a=0,c=a.
∴a=b=c.
∴△ABC為等邊三角形.