设实数x,y,z都不等于1,且xyz=1.试证
1/(x-1)^2+1/(y-1)^2+1/(z-1)^2≥1

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设实数x,y,z都不等于1,且xyz=1.试证
1/(x-1)^2+1/(y-1)^2+1/(z-1)^2≥1 (1)

证明 设a,b,c是不相等的实数,因为xyz=1,那么
令x=(c-b)/(a-c),y=(a-c)/(b-a),z=(b-a)/(c-b).
对(1)式左边置换得:
T=1/(x-1)^2+1/(y-1)^2+1/(z-1)^2=
(a-c)^2/(2c-a-b)^2+(b-a)^2/(2a-b-c)^2+(c-b)^2/(2b-c-a)^2.

再设a-b=m, n=b-c,则a-c=m+n.[m,n∈R].
T=(m+n)^2/(m+2n)^2+m^2/(2m+n)^2+n^2/(m-n)^2.

故(1)式等价于
(m+n)^2/(m+2n)^2+m^2/(2m+n)^2+n^2/(m-n)^2≥1 (2)

(2)式去分母化简为
m^6-3n*m^5-6n^2*m^4+34n^3*m^3+42n^4*m^2+12n^5*m+n^6≥0

<==> (m^3-3n*m^2-6n^2*m-n^3)^2≥0.
上式显然成立.