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Given-a-b-c-real-and-positive-numbers-and-a-b-c-1-Find-the-minimum-value-of-a-b-abc-




Question Number 22154 by Joel577 last updated on 12/Oct/17
Given a,b,c real and positive numbers, and  a + b + c = 1  Find the minimum value of  ((a + b)/(abc))
Givena,b,crealandpositivenumbers,anda+b+c=1Findtheminimumvalueofa+babc
Answered by ajfour last updated on 12/Oct/17
let a=cx    and  b=cy  then   c(x+y+1)=1  z=((a+b)/(abc)) =((x+y)/(c^2 xy)) =(((x+y)(x+y+1)^2 )/(xy))  As z is minimum,  (∂z/∂x) =((x[(x+y+1)^2 +2(x+y)(x+y+1)]−(x+y)(x+y+1)^2 )/(x^2 y)) =0  ⇒ x[x+y+1+2x+2y]−(x+y)(x+y+1)=0  3x^2 +3xy+x−x^2 −y^2 −2xy−x−y=0  2x^2 +xy−y^2 −y=0  2x^2 +(x−1)y−y^2 =0       ...(i)  similarly  (∂z/∂y)=0  shall give  2y^2 +(y−1)x−x^2 =0     ...(ii)  ⇒   x=y  so   2x^2 +x^2 −x−x^2 =0  ⇒      2x^2 −x=0         or     x(x−(1/2))=0    we shall choose x=y=(1/2) since  a, b, and c are +ve real numbers.  Then    a=(c/2)    ;    b=(c/2)        ⇒   (c/2)+(c/2)+c =1     so     c=(1/2) ,  a=(1/4) ,   b=(1/4)   ⇒  min( ((a+b)/(abc)) )= ((1/2)/(1/32)) =16 .
leta=cxandb=cythenc(x+y+1)=1z=a+babc=x+yc2xy=(x+y)(x+y+1)2xyAszisminimum,zx=x[(x+y+1)2+2(x+y)(x+y+1)](x+y)(x+y+1)2x2y=0x[x+y+1+2x+2y](x+y)(x+y+1)=03x2+3xy+xx2y22xyxy=02x2+xyy2y=02x2+(x1)yy2=0(i)similarlyzy=0shallgive2y2+(y1)xx2=0(ii)x=yso2x2+x2xx2=02x2x=0orx(x12)=0weshallchoosex=y=12sincea,b,andcare+verealnumbers.Thena=c2;b=c2c2+c2+c=1soc=12,a=14,b=14min(a+babc)=1/21/32=16.
Commented by Joel577 last updated on 12/Oct/17
Thank you very much Sir
ThankyouverymuchSir
Commented by Joel577 last updated on 12/Oct/17
I have heard about inequality AM ≥ GM  is it possible to use that inequality?
IhaveheardaboutinequalityAMGMisitpossibletousethatinequality?
Answered by ajfour last updated on 12/Oct/17
let  a+b=p  and   f=((a+b)/(abc)) =(p/(abc))   = ((4p)/((4ab)c))   ⇒      f =((4p)/([(a+b)^2 −(a−b)^2 ]c))       or  f=((4p)/([p^2 −(a−b)^2 ]c))  For f to be minimum ,   a=b  ⇒     f=((4p)/(p^2 c)) = (4/(pc))       but a+b+c =1  ⇒   p+c=1  max. of cp results when c=p  ⇒     c=(1/2) and p=(1/2)         So  f =(4/(cp)) =(4/((1/4))) =16 .
leta+b=pandf=a+babc=pabc=4p(4ab)cf=4p[(a+b)2(ab)2]corf=4p[p2(ab)2]cForftobeminimum,a=bf=4pp2c=4pcbuta+b+c=1p+c=1max.ofcpresultswhenc=pc=12andp=12Sof=4cp=4(1/4)=16.
Commented by Joel577 last updated on 13/Oct/17
thank you very much
thankyouverymuch

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