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Question

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 B 24
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[x34(yzx)4(yzx)4(yzx)8(1yz)3/28(1yz)3/2]1/6
f(x)24
Minimum value of f(x) is 24.

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