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x+y+z=1,F=2x2+y2+3z2的最小值为()A.611B.713C.58D.49
题目详情
x+y+z=1,F=2x2+y2+3z2的最小值为( )
A.
B.
C.
D.
A.
6 |
11 |
B.
7 |
13 |
C.
5 |
8 |
D.
4 |
9 |
▼优质解答
答案和解析
由柯西不等式得,(2x2+y2+3z2)(
+1+
)≥(2x2×
+y2×1+3z2×
)22=(x+y+z)2=1
∴2x2+y2+3z2≥
,即F的最小值为
故选A.
1 |
2 |
1 |
3 |
1 |
2 |
1 |
3 |
∴2x2+y2+3z2≥
6 |
11 |
6 |
11 |
故选A.
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