Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

$$\mathrm{i}\:\mathrm{just}\:\mathrm{used}\:\mathrm{AM}-\mathrm{GM}\:\mathrm{inequality}\:\mathrm{which} \\ $$$$\mathrm{is}\:\mathrm{true}\:\mathrm{for}\:\mathrm{all}\:\mathrm{real}\:\mathrm{numbers} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com