what-s-the-minimum-value-of-a-1-b-a-b-where-a-gt-b-gt-0-a-b-R-

Question Number 189803 by uchihayahia last updated on 22/Mar/23

{w h a t}^{'} s t h e m i n i m u m v a l u e o f

a + \frac{1}{b (a - b)} w h e r e a > b > 0 a, b \in R

Answered by cortano12 last updated on 22/Mar/23

f (a, b) = a + b^{- 1} {(a - b)}^{- 1} = a + {(a b - b^{2})}^{- 1}

\frac{\partial f}{\partial a} = 1 - b {(a b - b^{2})}^{- 2} = 0

\frac{\partial f}{\partial b} = - (a - 2 b) {(a b - b^{2})}^{- 2} = 0

{\begin{cases} 1 = \frac{b}{{(a b - b^{2})}^{2}} \Rightarrow 1 = \frac{1}{b {(2 b - b)}^{2}} \\ \frac{2 b - a}{{(a b - b^{2})}^{2}} = 0 \Rightarrow a = 2 b \end{cases}

\Rightarrow b^{3} = 1 \Rightarrow {\begin{cases} b = 1 \\ a = 2 \end{cases}

f (2, 1) = 2 + \frac{1}{1 . (2 - 1)} = 3

Commented by uchihayahia last updated on 23/Mar/23

t h a n k y o u

Answered by mr W last updated on 22/Mar/23

a + \frac{1}{b (a - b)}

= a - b + b + \frac{1}{b (a - b)}

= c + b + \frac{1}{b c} (w i t h c = a - b > 0)

⩾ 3 \sqrt[3]{c \times b \times \frac{1}{b c}} = 3

\Rightarrow m i n i m u m i s 3

Commented by manxsol last updated on 22/Mar/23

t h a n k s, S i r W y S r . C o r t a n o

Commented by mehdee42 last updated on 22/Mar/23

B r a v o . V e r y b e a u t i f u l

Commented by uchihayahia last updated on 23/Mar/23

t h a n k s t h i s i s w h a t i l o o k i n g f o r

Answered by ajfour last updated on 23/Mar/23

f = \frac{a b}{b} + \frac{1}{a b - b^{2}}

= \frac{1}{\sqrt{b}} (\frac{a b - b^{2}}{\sqrt{b}} + \frac{\sqrt{b}}{a b - b^{2}}) + b

= \frac{2}{\sqrt{b}} + b

\frac{d f}{d b} = - \frac{1}{b \sqrt{b}} + 1 = 0 \Rightarrow b = 1

f_{m i n} = 2 + 1

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