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| xyz inequality (Posted on 2015-07-24) |
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If x, y, and z are nonnegative real numbers,
then prove that
x2 + xy2 + xyz2 ≥ 4xyz - 4
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Submitted by Bractals
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Solution:
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x2 + xy2 + xyz2 ≥ 4xyz - 4
if and only if
(x - 2)2 + x(y - 2)2 + xy(z - 2)2 ≥ 0
QED
Note: the inequality is true even if z is any
real number instead of nonnegative.
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Subject |
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Date |
 | solution | Daniel | 2015-07-24 08:45:54 |
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