if-a-gt-b-gt-0-find-the-minimum-of-a-2-1-a-b-b-

Question Number 190192 by mr W last updated on 29/Mar/23

i f a > b > 0, f i n d t h e m i n i m u m o f

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

Commented by Frix last updated on 29/Mar/23

Minimum = 4 at a = \sqrt{2} \land b = \frac{\sqrt{2}}{2}

Commented by mr W last updated on 29/Mar/23

{w h a t}^{'} s y o u r s o l u t i o n ?

Answered by cortano12 last updated on 29/Mar/23

z = a^{2} + {(ab - b^{2})}^{- 1}

\frac{dz}{da} = 2 a - b {(ab - b^{2})}^{- 2}

\frac{dz}{db} = - (a - 2 b) {(ab - b^{2})}^{- 2}

\frac{dz}{da} = 0 \Rightarrow 2 a = \frac{b}{b^{2} {(a - b)}^{2}}

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

\frac{dz}{db} = 0 \Rightarrow \frac{2 b - a}{{(ab - b^{2})}^{2}} = 0; a = 2 b (ii)

\Rightarrow 4 b = \frac{1}{b^{3}}; b = \frac{1}{\sqrt{2}} \land a = \sqrt{2}

\Rightarrow z_{\min} = 2 + \frac{\sqrt{2}}{(\sqrt{2} - \frac{1}{2} \sqrt{2})} = 4

Commented by mr W last updated on 29/Mar/23

t h a n k s!

Answered by witcher3 last updated on 29/Mar/23

(a - b) b ⩽ \frac{1}{4} {((a - b) + b)}^{2} \dots AM, GM

\Rightarrow \frac{1}{b (a - b)} ⩾ \frac{4}{a^{2}}

a^{2} + \frac{1}{(a - b) b} ⩾ a^{2} + \frac{4}{a^{2}} ⩾ 2 \sqrt{. a^{2} . \frac{4}{a^{2}}} = 4 \dots AM, GM

Commented by mr W last updated on 29/Mar/23

t h a n k s!

Answered by mr W last updated on 29/Mar/23

s a y c = a - b > 0

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

= {(c + b)}^{2} + \frac{1}{c b}

= c^{2} + b^{2} + 2 b c + \frac{1}{c b}

= c^{2} + b^{2} + b c + b c + \frac{1}{4 c b} + \frac{1}{4 c b} + \frac{1}{4 c b} + \frac{1}{4 c b}

⩾ 8 \sqrt[8]{c^{2} \times b^{2} \times {(b c)}^{2} \times {(\frac{1}{4 c b})}^{4}} = 8 \times \frac{1}{2} = 4

\Rightarrow m i n i m u m = 4

w h e n c^{2} = b^{2} = b c = \frac{1}{4 c b}, i . e . c = b = \frac{1}{\sqrt{2}} & a = \sqrt{2}

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