Question

已知方程组x+y+z=0 x^2+y^2+z^2=1,求

x+y+z=a,x^2+y^2+z^2=1/2a^2,求x,y,z取值范围
已知x,y,z均为实数,a>0,且满足x+y+z=a,x^2+y^2+z^2=1/2 a^2,
求x,y,z的取值范围.

Answer 1

2(a^2/2-x^2)-(a-x)^2=2(y^2+z^2)-(y+z)^2=(y-z)^2>=0
整理得
(2a-3x)x>=0
解得
0