题目内容
若不等式
+
≤k
对于任意正实数x、y成立,则k的取值范围为
______.
| x |
| y |
| 2x+y |
显然k>0,故k2≥
.
令t=
>0,则k2≥
=
(1+
)
令u=4t+1>1,则t=
.
可转化为:s(u)=
=
≤2,
于是,
(1+
)≤
(1+2)=
.
∴k2≥
,即k≥
时,不等式恒成立(当x=4y>0时等号成立).
故答案为:[
,+∞)
x+y+2
| ||
| 2x+y |
令t=
|
| y(t2+2t+1) |
| y(2t2+1) |
| 1 |
| 2 |
| 4t+1 |
| 2t2+1 |
令u=4t+1>1,则t=
| u-1 |
| 4 |
| 4t+1 |
| 2t2+1 |
| 8u |
| u2-2u+9 |
| 8 | ||
u+
|
于是,
| 1 |
| 2 |
| 4t+1 |
| 2t2+1 |
| 1 |
| 2 |
| 3 |
| 2 |
∴k2≥
| 3 |
| 2 |
| ||
| 2 |
故答案为:[
| ||
| 2 |
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