已知正数x,y,z满足x+2y+3z=1,则1x+2y
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已知正数x,y,z满足x+2y+3z=1,则
1 |
x+2y |
由柯西不等式可得(x+2y+2y+3z+3z+x)(
1
x+2y+
4
2y+3z+
9
3z+x)≥(1+2+3)2,
∵x+2y+3z=1,
∴2(
1
x+2y+
4
2y+3z+
9
3z+x)≥36,
∴
1
x+2y+
4
2y+3z+
9
3z+x≥18,
∴
1
x+2y+
4
2y+3z+
9
3z+x的最小值为18.
故答案为:18.
1
x+2y+
4
2y+3z+
9
3z+x)≥(1+2+3)2,
∵x+2y+3z=1,
∴2(
1
x+2y+
4
2y+3z+
9
3z+x)≥36,
∴
1
x+2y+
4
2y+3z+
9
3z+x≥18,
∴
1
x+2y+
4
2y+3z+
9
3z+x的最小值为18.
故答案为:18.
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