本试题主要是考查了不等式中绝对值不等式的恒成立问题的运用,以及均值不等式的综合求解。,根据>0,b>0 且a+b="1" ,对于
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232241165091217.png)
,下一步就是求解绝对值不等式即可。
解:∵ a>0,b>0 且a+b="1" ∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823224116525327.png)
+
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823224116556387.png)
="(a+b)("
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823224116525327.png)
+
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823224116556387.png)
)=5+
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823224116728396.png)
+
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823224116743473.png)
≥9,
故
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823224116525327.png)
+
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823224116556387.png)
的最小值为9, ------------------------5分
因为对a,b∈(0,+∞),使
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823224116525327.png)
+
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823224116556387.png)
≥|2x-1|-|x+1|恒成立,所以,|2x-1|-|x+1|≤9, -7分
当 x≤-1时,2-x≤9, ∴ -7≤x≤-1, 当 -1<x<
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823224116915338.png)
时,-3x≤9,
∴ -1<x<
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823224116915338.png)
,当 x≥
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823224116915338.png)
时,x-2≤9, ∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823224116915338.png)
≤x≤11,∴ -7≤x≤11 ------------- 10分