题目内容
（1）设x＞y＞z，n∈R*，且
+
≥
恒成立，求n的最大值．
（2）已知函数f（x）=2x的反函数是f-1（x），若f-1（a）+f-1（b）=4（a，b∈R*），求
+
的最小值．
| 1 |
| x-y |
| 1 |
| y-z |
| n |
| x-z |
（2）已知函数f（x）=2x的反函数是f-1（x），若f-1（a）+f-1（b）=4（a，b∈R*），求
| 1 |
| a |
| 4 |
| b |
分析：（1）由x＞y＞z，n∈R*原不等式可等价变形为n≤
+
再配凑为
+
=
+
，利用基本不等式求最值．
（2）反函数f-1(x)=
．f-1(a)+f-1(b)=
+
=4，a+b=8．从而，
+
=
(a+b)(
+
)=
(1+4+
+
)≥
(5+2
)=
．
| x-z |
| x-y |
| x-z |
| y-z |
| x-z |
| x-y |
| x-z |
| y-z |
| x-y+y-z |
| x-y |
| x-y+y-z |
| y-z |
（2）反函数f-1(x)=
| x |
| 2 |
| a |
| 2 |
| b |
| 2 |
| 1 |
| a |
| 4 |
| b |
| 1 |
| 8 |
| 1 |
| a |
| 4 |
| b |
| 1 |
| 8 |
| b |
| a |
| 4a |
| b |
| 1 |
| 8 |
|
| 9 |
| 8 |
解答：解：（1）∵x＞y＞z，n∈R*∴原不等式可等价变形为n≤
+
+
=
+
=1+
+
≥2
+2=4，∴n不能大于
+
的最小值4，∴n的最大值是4．
（2）由函数f（x）=2x可得，反函数f-1(x)=
．
∴f-1(a)+f-1(b)=
+
=4，a+b=8．从而，
+
=
(a+b)(
+
)=
(1+4+
+
)≥
(5+2
)=
．
| x-z |
| x-y |
| x-z |
| y-z |
| x-z |
| x-y |
| x-z |
| y-z |
| x-y+y-z |
| x-y |
| x-y+y-z |
| y-z |
| y-z |
| x-y |
| x-y |
| y-z |
|
| x-z |
| x-y |
| x-z |
| y-z |
（2）由函数f（x）=2x可得，反函数f-1(x)=
| x |
| 2 |
∴f-1(a)+f-1(b)=
| a |
| 2 |
| b |
| 2 |
| 1 |
| a |
| 4 |
| b |
| 1 |
| 8 |
| 1 |
| a |
| 4 |
| b |
| 1 |
| 8 |
| b |
| a |
| 4a |
| b |
| 1 |
| 8 |
|
| 9 |
| 8 |
点评：本题考查基本不等式的应用：求最值．基本不等式求最值时要注意三个原则：一正，即各项的取值为正；二定，即各项的和或积为定值；三相等，即要保证取等号的条件成立
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