已知x1,x2,x3,x4为实数,且x1+x2+x3+x4=6,x1^2+x2^2+x3^2+x4^2=12,求证:0=

x1=6-(x2+x3+x4)假设x2+x3+x4=m 代入x1+x2+x3+x4=6,x1^2+x2^2+x3^2+x4^2=12得
(6-m)^2+x2^2+x3^2+x4^2=12>=(6-m)^2+m^2/3得3<=m<=6
所以0<=x1<=3
同理可得：0<=x2<=3,.0<=xi<=3.(i=1.2.3.4)