（关于恒等式 a^2+b^2+c^2-ab-bc-ac=2/1

因为x-y=a z-y=10
两者相减x-z=a-10
因为x^2+y^2+z^2-xy-yz-zx
=1/2[2x^2+2y^2+2z^2-2xy-2yz-2zx]
=1/2[(x-y)^2+(x-z)^2+(y-z)^2]
把前三个等式代入原式
原式=1/2[a^2+(a-10)^2+(-10)^2]
=1/2[2a^2-20a+200]
=a^2-10a+100
=(a-5)^2+75
当a为5时,原式有最小值75
x+y+z=a^2+b^2+c^2-ab-bc-ac=1/2 [...]>=0

1.x^2+y^2+z^2-xy-yz-zx=(a-10)^2+75
2.x+y+x=1/2[(a-b)^2+(b-c)^2+(c-a)^2]>0(设a、b、c是不全相等的实数)
所以求证x、y、z至少有一个大于零

x^2+y^2+z^2-xy-yz-zx
=1/2 [(X-Y)^2+(Y-Z)^2+(Z-X)^2]
=1/2 [a^2+100+(10-a)^2]
=a^2-10a+100
=(a-5)^2+75>=75
x+y+z=a^2+b^2+c^2-ab-bc-ac=1/2 [...]>=0