a+b+c=x^2-2x+y^2-2y+z^2-2z+7π/6
=(x-1)^2+(y-1)^2+(z-1)^2+7π/6-3
因为π>3
所以7π/6>3
所以(x-1)^2+(y-1)^2+(z-1)^2+7π/6-3>0
所以abc至少有一个大于0
a+b+c=x^2-2x+y^2-2y+z^2-2z+7π/6
=(x-1)^2+(y-1)^2+(z-1)^2+7π/6-3
因为π>3
所以7π/6>3
所以(x-1)^2+(y-1)^2+(z-1)^2+7π/6-3>0
所以abc至少有一个大于0