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Question Number 110951 by ZiYangLee last updated on 01/Sep/20

	Answered by $@y@m last updated on 01/Sep/20
	$Let y=kx (0<k<1) Let z=x+(1/(y(x−y))) ⇒ z=x+(1/(kx(x−kx))) ⇒z =x+(1/(k(1−k)x^2 )) .....(1) (dz/dx)=1−(2/(k(1−k)x^3 )) ....(2) For maxima or minima, (dz/dx)=0 1−(2/(k(1−k)x^3 )) =0 1=(2/(k(1−k)x^3 )) x^3 =(2/(k(1−k))) x={(2/(k(1−k)))}^(1/3) ...(3) (d^2 z/dx^2 )=(6/(k(1−k)x^4 )) (d^2 z/dx^2 )]_(at x={(2/(k(1−k)))}^(1/3) ) =(6/(k(1−k){(2/(k(1−k)))}^(4/3) )) =((6{k(1−k)}^(1/3) )/2^(4/3) )>0 (∵ 0<k<1) ∴ z is minimum when x={(2/(k(1−k)))}^(1/3) From (1), z =x+(1/(k(1−k)x^2 )) ⇒z=((k(1−k)x^3 +1)/(k(1−k)x^2 )) ⇒z_(min.) =((k(1−k)×(2/(k(1−k)))+1)/(k(1−k){(2/(k(1−k)))}^(2/3) )) ⇒z_(min.) =((3{k(1−k)}^(1/3) )/2^(2/3) ) ....(4) Now, let v=k(1−k) It can be shown that v is maximum when k=(1/2). v_(max.) =(1/4) From (4), it is evident that z is minimum when v is maximum. =3.((1/2))^(2/3) ((1/2))^(2/3) =3.((2/3))^(4/3)$
	$L e t y = k x (0 < k < 1)$ $L e t z = x + \frac{1}{y (x - y)}$ $\Rightarrow z = x + \frac{1}{k x (x - k x)}$ $\Rightarrow z = x + \frac{1}{k (1 - k) x^{2}} . . . . . (1)$ $\frac{d z}{d x} = 1 - \frac{2}{k (1 - k) x^{3}} . . . . (2)$ $F o r m a x i m a o r m i n i m a,$ $\frac{d z}{d x} = 0$ $1 - \frac{2}{k (1 - k) x^{3}} = 0$ $1 = \frac{2}{k (1 - k) x^{3}}$ $x^{3} = \frac{2}{k (1 - k)}$ $x = {\frac{2}{k (1 - k)}}^{\frac{1}{3}} . . . (3)$ $\frac{d^{2} z}{{d x}^{2}} = \frac{6}{k (1 - k) x^{4}}$ ${\frac{d^{2} z}{{d x}^{2}}]}_{a t x = {\frac{2}{k (1 - k)}}^{\frac{1}{3}}} = \frac{6}{k (1 - k) {\frac{2}{k (1 - k)}}^{\frac{4}{3}}}$ $= \frac{6 {k (1 - k)}^{\frac{1}{3}}}{2^{\frac{4}{3}}} > 0 (∵ 0 < k < 1)$ $∴ z i s m i n i m u m w h e n x = {\frac{2}{k (1 - k)}}^{\frac{1}{3}}$ $F r o m (1),$ $z = x + \frac{1}{k (1 - k) x^{2}}$ $\Rightarrow z = \frac{k (1 - k) x^{3} + 1}{k (1 - k) x^{2}}$ $\Rightarrow z_{m i n .} = \frac{k (1 - k) \times \frac{2}{k (1 - k)} + 1}{k (1 - k) {\frac{2}{k (1 - k)}}^{\frac{2}{3}}}$ $\Rightarrow z_{m i n .} = \frac{3 {k (1 - k)}^{\frac{1}{3}}}{2^{\frac{2}{3}}} . . . . (4)$ $N o w, l e t v = k (1 - k)$ $I t c a n b e s h o w n t h a t v i s m a x i m u m$ $w h e n k = \frac{1}{2} .$ $v_{m a x .} = \frac{1}{4}$ $F r o m (4), i t i s e v i d e n t t h a t z i s m i n i m u m w h e n v i s m a x i m u m .$ $= 3 . {(\frac{1}{2})}^{\frac{2}{3}} {(\frac{1}{2})}^{\frac{2}{3}}$ $= 3 . {(\frac{2}{3})}^{\frac{4}{3}}$
	Commented by ZiYangLee last updated on 01/Sep/20

	$Thanks Bro!$
	Commented by $@y@m last updated on 01/Sep/20

	$I s t h e a n s w e r c o r r e c t ?$ $I a m n o t s u r e a b o u t t h e s o l u t i o n .$
	Answered by 1549442205PVT last updated on 01/Sep/20
	$set x−y=m>0;P=m+y+(1/(my)) ≥3^3 (√(m.y.(1/(my))))=3 The equality ocurrs if and only if m=y=(1/(my))⇔ { ((m^2 y=1)),((my^2 =1)) :}⇒my=1⇒m=y=1 x=y+1 Thus,P=x+(1/(y(x−y))) has smallest value equal to 3 when x=2,y=1$
	$set x - y = m > 0; P = m + y + \frac{1}{my}$ $⩾ 3^{3} \sqrt{m . y . \frac{1}{my}} = 3$ $The equality ocurrs if and only if$ $m = y = \frac{1}{my} \Leftrightarrow {\begin{cases} m^{2} y = 1 \\ {my}^{2} = 1 \end{cases} \Rightarrow my = 1 \Rightarrow m = y = 1$ $x = y + 1$ $Thus, P = x + \frac{1}{y (x - y)} has smallest$ $value equal to 3 when x = 2, y = 1$
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