已知a,b,c属于R,a^2+b^2+c^2=1.求证,|a+b+c|=（a+b+c)^2对满足题条件的实数a,b,c恒

(1) (a+b+c)²=a²+b²+c²+2ab+2bc+2ca
≤a²+b²+c²+a²+b²+b²+c²+c²+a²
=3(a²+b²+c²)=3
所以 |a+b+c|≤√3
(2)|x-1|+|x+1|≥(a+b+c)²,
从而 |x-1|+|x+1|≥[(a+b+c)²]max
即 |x-1|+|x+1|≥3
故只需 |x+1+x-1|≥3
|x|≥3/2
x≥3/2或x≤-3/2.