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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-

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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.
Giventhatx,y,zarerealnumberssuchthatx+y+z=0andxyz=−432.Ifa=1x+1y+1z,findthesmallestpossiblevalueofa.
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
Sincex+y+z=0,amongoftheenumbersthereexistsleastatonepositivenumberWLOGsupposez>0a=1x+1y+1z=xy+yz+zxxyz=xy+yz+zx−432=(x+y+z)2−(x2+y2+z2)−432.2=x2+y2+z2864=x2+y2+(x+y)2864=2(x+y)2−2xy864=z2+432z432=z2+216z+216z432⩾33z2.216z.216z/432=3.6236.12=14Theequalityocurrsifandonlyif{z2=216zx+y+z=0xyz=−432⇔{z=6x=6y=−12Thus,amin=14when(x,y,6)=(6,−12,6)andallitspermutation
Commented by ZiYangLee last updated on 16/Oct/20
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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)
⇒a=x+yxy−1x+y=(x+y)2−xyxy(x+y)⇒a=(x−y)2+3xy432,aminwhenx−y=0orx=y.itfollowsthatz=−2xandx2(−2x)=−432⇒x3=216,wegetx=6=yandz=−12.Thusminimumvalueofa=16+16−112=412−112=14

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