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Question Number 109595 by nimnim last updated on 24/Aug/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)
$${For}\:{any}\:{Real}\:{numbers}\:{x},{y}\:{and}\:{z}, \\ $$$$\:{if}\:\:\left({x}+{y}+{z}\right)=\mathrm{2},\:{then}\:{prove}\:{that} \\ $$$$\:\:\:\:\:\:\:\:{xyz}\geqslant\mathrm{8}\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{y}\right)\left(\mathrm{1}−{z}\right) \\ $$
Commented by 1549442205PVT last updated on 25/Aug/20
This inequality is only true for x,y,z>0 .Example,for x=−(2/3),y=(6/5) z=((22)/(15)) then it is false.Then xyz−8(1−x)(1−y)(1−z)=−((544)/(25))<0
$$\mathrm{This}\:\mathrm{inequality}\:\mathrm{is}\:\mathrm{only}\:\mathrm{true}\:\mathrm{for} \\ $$$$\mathrm{x},\mathrm{y},\mathrm{z}>\mathrm{0}\:.\mathrm{Example},\mathrm{for}\:\mathrm{x}=−\frac{\mathrm{2}}{\mathrm{3}},\mathrm{y}=\frac{\mathrm{6}}{\mathrm{5}} \\ $$$$\mathrm{z}=\frac{\mathrm{22}}{\mathrm{15}}\:\mathrm{then}\:\mathrm{it}\:\mathrm{is}\:\mathrm{false}.\mathrm{Then} \\ $$$$\boldsymbol{\mathrm{xyz}}−\mathrm{8}\left(\mathrm{1}−\boldsymbol{\mathrm{x}}\right)\left(\mathrm{1}−\boldsymbol{\mathrm{y}}\right)\left(\mathrm{1}−\boldsymbol{\mathrm{z}}\right)=−\frac{\mathrm{544}}{\mathrm{25}}<\mathrm{0} \\ $$
Answered by floor(10²Eta[1]) last updated on 24/Aug/20
((x+y+z)/3)≥^3 (√(xyz))⇒(2/3)≥^3 (√(xyz))∴(8/(27))≥xyz ⇒xyz≥8(1−x)(1−y)(1−z) ⇔(1/(27))≥(1−x)(1−y)(1−z) (((1−x)+(1−y)+(1−z))/3)≥^3 (√((1−x)(1−y)(1−z))) ((3−(x+y+z))/3)≥^3 (√((1−x)(1−y)(1−z))) (1/3)≥^3 (√((1−x)(1−y)(1−z))) (1/(27))≥(1−x)(1−y)(1−z) ∴xyz≥8(1−x)(1−y)(1−z), x+y+z=2
$$\frac{\mathrm{x}+\mathrm{y}+\mathrm{z}}{\mathrm{3}}\geqslant^{\mathrm{3}} \sqrt{\mathrm{xyz}}\Rightarrow\frac{\mathrm{2}}{\mathrm{3}}\geqslant^{\mathrm{3}} \sqrt{\mathrm{xyz}}\therefore\frac{\mathrm{8}}{\mathrm{27}}\geqslant\mathrm{xyz} \\ $$$$\Rightarrow\mathrm{xyz}\geqslant\mathrm{8}\left(\mathrm{1}−\mathrm{x}\right)\left(\mathrm{1}−\mathrm{y}\right)\left(\mathrm{1}−\mathrm{z}\right) \\ $$$$\Leftrightarrow\frac{\mathrm{1}}{\mathrm{27}}\geqslant\left(\mathrm{1}−\mathrm{x}\right)\left(\mathrm{1}−\mathrm{y}\right)\left(\mathrm{1}−\mathrm{z}\right) \\ $$$$\frac{\left(\mathrm{1}−\mathrm{x}\right)+\left(\mathrm{1}−\mathrm{y}\right)+\left(\mathrm{1}−\mathrm{z}\right)}{\mathrm{3}}\geqslant^{\mathrm{3}} \sqrt{\left(\mathrm{1}−\mathrm{x}\right)\left(\mathrm{1}−\mathrm{y}\right)\left(\mathrm{1}−\mathrm{z}\right)} \\ $$$$\frac{\mathrm{3}−\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)}{\mathrm{3}}\geqslant^{\mathrm{3}} \sqrt{\left(\mathrm{1}−\mathrm{x}\right)\left(\mathrm{1}−\mathrm{y}\right)\left(\mathrm{1}−\mathrm{z}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\geqslant^{\mathrm{3}} \sqrt{\left(\mathrm{1}−\mathrm{x}\right)\left(\mathrm{1}−\mathrm{y}\right)\left(\mathrm{1}−\mathrm{z}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{27}}\geqslant\left(\mathrm{1}−\mathrm{x}\right)\left(\mathrm{1}−\mathrm{y}\right)\left(\mathrm{1}−\mathrm{z}\right)\: \\ $$$$\therefore\mathrm{xyz}\geqslant\mathrm{8}\left(\mathrm{1}−\mathrm{x}\right)\left(\mathrm{1}−\mathrm{y}\right)\left(\mathrm{1}−\mathrm{z}\right),\:\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{2} \\ $$
Commented by nimnim last updated on 24/Aug/20
Thank you sir..
$${Thank}\:{you}\:{sir}.. \\ $$
Commented by floor(10²Eta[1]) last updated on 25/Aug/20
i just used AM-GM inequality which is true for all real numbers
$$\mathrm{i}\:\mathrm{just}\:\mathrm{used}\:\mathrm{AM}-\mathrm{GM}\:\mathrm{inequality}\:\mathrm{which} \\ $$$$\mathrm{is}\:\mathrm{true}\:\mathrm{for}\:\mathrm{all}\:\mathrm{real}\:\mathrm{numbers} \\ $$

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