设{a=x+2y+z,b=x+y+2z,c=x+y+3z}
则{a-b=y-z,c-b=z}
∴{x+3z=2b-a,y=c+a-2b,z=c-b}
从而,
(x+3z)/(x+2y+z)+4y/(x+y+2z)-8z/(x+y+3z)
=(2b-a)/a+4(c+a-2b)/b-8(c-b)/c
=-17+(2b/a+4a/b)+(4c/b+8b/c)
≥-17+2根(2b/a*4a/b)+2根(4c/b*8b/c)
=-17+12根2.
取x=3-2根2,y=-1+根2,z=根2时等号成立.
∴f(x,y,z)|min=-17+12根2.
答：平方在利用均值不等式详情>>
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