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Question Number 118239 by ZiYangLee last updated on 16/Oct/20
Given that x,y,z are real numbers such that x+y+z=0 and xyz=−432. If a=(1/x)+(1/y)+(1/z), find the smallest possible value of a.
$$\mathrm{Given}\:\mathrm{that}\:{x},{y},{z}\:\mathrm{are}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{such}\:\mathrm{that} \\ $$$${x}+{y}+{z}=\mathrm{0}\:\mathrm{and}\:{xyz}=−\mathrm{432}. \\ $$$$\mathrm{If}\:{a}=\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}},\:\mathrm{find}\:\mathrm{the}\:\mathrm{smallest}\:\mathrm{possible} \\ $$$$\mathrm{value}\:\mathrm{of}\:{a}. \\ $$
Answered by 1549442205PVT last updated on 16/Oct/20
Since x+y+z=0,among of thee numbers there exists least at one positive number WLOG suppose z>0 a=(1/x)+(1/y)+(1/z)=((xy+yz+zx)/(xyz))=((xy+yz+zx)/(−432)) =(((x+y+z)^2 −(x^2 +y^2 +z^2 ))/(−432.2))=((x^2 +y^2 +z^2 )/(864)) =((x^2 +y^2 +(x+y)^2 )/(864))=((2(x+y)^2 −2xy)/(864)) =((z^2 +((432)/z))/(432))=((z^2 +((216)/z)+((216)/z))/(432)) ≥3^3 (√(z^2 .((216)/z).((216)/z)))/432=((3.6^2 )/(36.12))=(1/4) The equality ocurrs if and only if { ((z^2 =((216)/z))),((x+y+z=0)),((xyz=−432)) :}⇔ { ((z=6)),((x=6)),((y=−12)) :} Thus,a_(min) =(1/4) when (x,y,6)=(6,−12,6) and all its permutation
$$\mathrm{Since}\:\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{0},\mathrm{among}\:\mathrm{of}\:\mathrm{thee}\:\mathrm{numbers} \\ $$$$\mathrm{there}\:\mathrm{exists}\:\mathrm{least}\:\mathrm{at}\:\mathrm{one}\:\mathrm{positive}\:\mathrm{number} \\ $$$$\mathrm{WLOG}\:\mathrm{suppose}\:\mathrm{z}>\mathrm{0} \\ $$$$\mathrm{a}=\frac{\mathrm{1}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{y}}+\frac{\mathrm{1}}{\mathrm{z}}=\frac{\mathrm{xy}+\mathrm{yz}+\mathrm{zx}}{\mathrm{xyz}}=\frac{\mathrm{xy}+\mathrm{yz}+\mathrm{zx}}{−\mathrm{432}} \\ $$$$=\frac{\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)^{\mathrm{2}} −\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} \right)}{−\mathrm{432}.\mathrm{2}}=\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} }{\mathrm{864}} \\ $$$$=\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} }{\mathrm{864}}=\frac{\mathrm{2}\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} −\mathrm{2xy}}{\mathrm{864}} \\ $$$$=\frac{\mathrm{z}^{\mathrm{2}} +\frac{\mathrm{432}}{\mathrm{z}}}{\mathrm{432}}=\frac{\mathrm{z}^{\mathrm{2}} +\frac{\mathrm{216}}{\mathrm{z}}+\frac{\mathrm{216}}{\mathrm{z}}}{\mathrm{432}} \\ $$$$\geqslant\mathrm{3}\:^{\mathrm{3}} \sqrt{\mathrm{z}^{\mathrm{2}} .\frac{\mathrm{216}}{\mathrm{z}}.\frac{\mathrm{216}}{\mathrm{z}}}/\mathrm{432}=\frac{\mathrm{3}.\mathrm{6}^{\mathrm{2}} }{\mathrm{36}.\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{The}\:\mathrm{equality}\:\mathrm{ocurrs}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if} \\ $$$$\begin{cases}{\mathrm{z}^{\mathrm{2}} =\frac{\mathrm{216}}{\mathrm{z}}}\\{\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{0}}\\{\mathrm{xyz}=−\mathrm{432}}\end{cases}\Leftrightarrow\begin{cases}{\mathrm{z}=\mathrm{6}}\\{\mathrm{x}=\mathrm{6}}\\{\mathrm{y}=−\mathrm{12}}\end{cases} \\ $$$$\mathrm{Thus},\mathrm{a}_{\mathrm{min}} =\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{when}\:\left(\mathrm{x},\mathrm{y},\mathrm{6}\right)=\left(\mathrm{6},−\mathrm{12},\mathrm{6}\right) \\ $$$$\mathrm{and}\:\mathrm{all}\:\mathrm{its}\:\mathrm{permutation} \\ $$$$ \\ $$
Commented by ZiYangLee last updated on 16/Oct/20
wow!
$$\mathrm{wow}! \\ $$
Answered by bobhans last updated on 16/Oct/20
⇒a = ((x+y)/(xy)) − (1/(x+y)) = (((x+y)^2 −xy)/(xy(x+y))) ⇒a = (((x−y)^2 +3xy)/(432)) , a_(min) when x−y=0 or x=y . it follows that z=−2x and x^2 (−2x)=−432 ⇒x^3 =216 , we get x = 6=y and z = −12. Thus minimum value of a = (1/6)+(1/6)−(1/(12)) = (4/(12))−(1/(12))=(1/4)
$$\Rightarrow{a}\:=\:\frac{{x}+{y}}{{xy}}\:−\:\frac{\mathrm{1}}{{x}+{y}}\:=\:\frac{\left({x}+{y}\right)^{\mathrm{2}} −{xy}}{{xy}\left({x}+{y}\right)} \\ $$$$\Rightarrow{a}\:=\:\frac{\left({x}−{y}\right)^{\mathrm{2}} +\mathrm{3}{xy}}{\mathrm{432}}\:,\:{a}_{{min}} \:{when}\:{x}−{y}=\mathrm{0} \\ $$$${or}\:{x}={y}\:.\:{it}\:{follows}\:{that}\:{z}=−\mathrm{2}{x}\:{and}\: \\ $$$${x}^{\mathrm{2}} \left(−\mathrm{2}{x}\right)=−\mathrm{432}\:\Rightarrow{x}^{\mathrm{3}} =\mathrm{216}\:,\:{we}\:{get}\:{x}\:=\:\mathrm{6}={y} \\ $$$${and}\:{z}\:=\:−\mathrm{12}.\:{Thus}\:{minimum}\:{value}\: \\ $$$${of}\:{a}\:=\:\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{12}}\:=\:\frac{\mathrm{4}}{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$

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