【解】2(x^2+y^2)- (x+y)^2=(x-y)^2≥0，所以(x+y)^2≤2(x^2+y^2)因为x+y =a-z, x^2+y^2=a^2/2-z^2,由(x+y)^2≤2(x^2+y^2)得(a-z)^2≤2(1/2*a^2-z^2)=a^2-2z^2整理得z(3z-2a)<=0所以0≤z≤(2/3)a同理可得0≤y≤2a/3,0≤z≤2a/3
