what′s the minimum value of a+(1/(b(a−b))) where a>b>0 a,b∈R


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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
	$f(a,b)=a+b^(−1) (a−b)^(−1) =a+(ab−b^2 )^(−1) (∂f/∂a) =1−b(ab−b^2 )^(−2) =0 (∂f/∂b) =−(a−2b)(ab−b^2 )^(−2) =0 { ((1=(b/((ab−b^2 )^2 ))⇒1=(1/(b(2b−b)^2 )))),((((2b−a)/((ab−b^2 )^2 ))=0⇒a=2b)) :} ⇒b^3 = 1⇒ { ((b=1)),((a=2)) :} f(2,1)=2+(1/(1.(2−1))) = 3$
	$f (a, b) = a + b^{- 1} {(a - b)}^{- 1} = a + {(a b - b^{2})}^{- 1}$ $\frac{\partial f}{\partial a} = 1 - b {(a b - b^{2})}^{- 2} = 0$ $\frac{\partial f}{\partial b} = - (a - 2 b) {(a b - b^{2})}^{- 2} = 0$ ${\begin{cases} 1 = \frac{b}{{(a b - b^{2})}^{2}} \Rightarrow 1 = \frac{1}{b {(2 b - b)}^{2}} \\ \frac{2 b - a}{{(a b - b^{2})}^{2}} = 0 \Rightarrow a = 2 b \end{cases}$ $\Rightarrow b^{3} = 1 \Rightarrow {\begin{cases} b = 1 \\ a = 2 \end{cases}$ $f (2, 1) = 2 + \frac{1}{1 . (2 - 1)} = 3$
	Commented byuchihayahia 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 bymanxsol 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 bymehdee42 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 byuchihayahia 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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