If xyz ≠ 0, is x (y + z) ≥ 0? : DS Archive
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# If xyz ≠ 0, is x (y + z) ≥ 0?

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If xyz ≠ 0, is x (y + z) ≥ 0? [#permalink]

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28 Nov 2005, 07:18
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If xyz ≠ 0, is x (y + z) ≥ 0?

(1) │y + z│ = │y│ + │z│
(2) │x + y│ = │x│ + │y│
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Tanmoi.

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Re: Absolute value DS 2 [#permalink]

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28 Nov 2005, 07:35
Tanmoi wrote:
If xyz ≠ 0, is x (y + z) ≥ 0?

(1) │y + z│ = │y│ + │z│
(2) │x + y│ = │x│ + │y│

With 1) Z and Y should be both positive or both negative.
But we dont know the value of X == insuff

With 2) X and Y should be both positive or both negative.
But if both are negative and Z is positive and greater than Y, then equation is not satisfied. == insuff

Taking both together, XYZ can be all positive and all negative. This will satisfy the equation.
Hence C
Re: Absolute value DS 2   [#permalink] 28 Nov 2005, 07:35
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