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Question Number 112796 by nimnim last updated on 09/Sep/20
For any real numbers x,y and z if  x+y+z=2, then prove that  xyz≥8(1−x)(1−y)(1−z).
$${For}\:{any}\:{real}\:{numbers}\:{x},{y}\:{and}\:{z}\:{if} \\ $$$${x}+{y}+{z}=\mathrm{2},\:{then}\:{prove}\:{that} \\ $$$${xyz}\geqslant\mathrm{8}\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{y}\right)\left(\mathrm{1}−{z}\right). \\ $$
Commented by nimnim last updated on 10/Sep/20
Yes sir..
$$\mathrm{Yes}\:\mathrm{sir}.. \\ $$
Commented by 1549442205PVT last updated on 10/Sep/20
The inequality is false when x=−1.4  y=0.9,z=2.5 because xyz=−3.15  8(1−x)(1−y)(1−z)=−72/25=−2.88>−3.15  You need must look at your question   It seems posted ago and was warned
$$\mathrm{The}\:\mathrm{inequality}\:\mathrm{is}\:\mathrm{false}\:\mathrm{when}\:\mathrm{x}=−\mathrm{1}.\mathrm{4} \\ $$$$\mathrm{y}=\mathrm{0}.\mathrm{9},\mathrm{z}=\mathrm{2}.\mathrm{5}\:\mathrm{because}\:\mathrm{xyz}=−\mathrm{3}.\mathrm{15} \\ $$$$\mathrm{8}\left(\mathrm{1}−\mathrm{x}\right)\left(\mathrm{1}−\mathrm{y}\right)\left(\mathrm{1}−\mathrm{z}\right)=−\mathrm{72}/\mathrm{25}=−\mathrm{2}.\mathrm{88}>−\mathrm{3}.\mathrm{15} \\ $$$$\boldsymbol{\mathrm{You}}\:\boldsymbol{\mathrm{need}}\:\boldsymbol{\mathrm{must}}\:\boldsymbol{\mathrm{look}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{your}}\:\boldsymbol{\mathrm{question}}\: \\ $$$$\boldsymbol{\mathrm{It}}\:\boldsymbol{\mathrm{seems}}\:\boldsymbol{\mathrm{posted}}\:\boldsymbol{\mathrm{ago}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{was}}\:\boldsymbol{\mathrm{warned}} \\ $$

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