Question Number 22154 by Joel577 last updated on 12/Oct/17

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 .](https://www.tinkutara.com/question/Q22155.png)
Commented by Joel577 last updated on 12/Oct/17

Commented by Joel577 last updated on 12/Oct/17

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 .](https://www.tinkutara.com/question/Q22171.png)
Commented by Joel577 last updated on 13/Oct/17
