If x,y and z are all positive, then the minimum value of f(x,y,z)=x3+12(yzx)+16(1yz)3/2 is
A
14
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B
24
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C
42
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D
3
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Solution
The correct option is B24 Let x3,4(yzx),4(yzx),4(yzx),8(1yz)3/2,8(1yz)3/2
be six positive numbers.
Applying A.M.≥G.M., ⇒f(x)6≥[x3⋅4(yzx)⋅4(yzx)⋅4(yzx)⋅8(1yz)3/2⋅8(1yz)3/2]1/6 ⇒f(x)≥24
Minimum value of f(x) is 24.