一数学题，帮帮忙拉，快解呀

x=y+8
(y+8)y+z^2=-16
(y+4)^2+z^2=0
y=-4
z=0
x=4
x+y+z=0

x－y＝8
x＝8+y
xy＋z^2＝－16
(8+y)*y+z^2=-16
8y+y^2+z^2=-16
(y^2+8y+16)+z^2=0
(y+4)^2+z^2=0
z=0
y+4=0
y=-4
x=8+y=8+(-4)=4
x＋y＋z=4+(-4)+0=0

z^2=-16-xy≥0,
xy≤-16(等号成立,当且仅当z=0)............(1)
又x-y=8,所以x≥0,y≤0.
x(-y)≤[x+(-y)]^2/4=16,
xy≥-16(等号成立,当且仅当x=-y)...........(2)
由(1)(2),xy=-16,x=-y,z=0,x+y+z=0.

有理数x，y，z满足x－y＝8，xy＋z2＝－16，求证x＋y＋z＝0
x－y＝8 x=8+y xy＋z^2=(8+y)y+z^2=y^2+8y+z^2=(y+4)^2+z^2-16=-16
(y+4)^2+z^2=0 y+4=0 z=0 y=-4
x=8+y=8-4=4
x＋y＋z=4-4+0=0