Given a,b are positive number satisfy a + b 1 Find minimum value of S know : S = ab +

Question by Kaya Renger

Kaya Renger Coordinator

22/09/2017 at 11:21

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Given a,b are positive number satisfy a + b $\le$ 1

Find minimum value of S know :

S = ab + $\dfrac{1}{ab}$

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Faded 28/01/2018 at 20:42

No :< , we can do this !

Denote $t = \frac{1}{a b}$

; $\frac{1}{t} = a b$

$t = \frac{1}{a b} \geq \frac{1}{{(\frac{a + b}{2})}^{2}} = \frac{1}{{(\frac{1}{2})}^{2}} = 4$

Predict a = b = $\frac{1}{2}$

then t = 4 , we have :

$S = \frac{1}{t} + t = (\frac{1}{t} + \frac{t}{16}) + \frac{15 t}{16} \geq 2. \sqrt{\frac{1}{t} . \frac{t}{16}} + \frac{15.4}{16} = \frac{17}{4}$

So $M i n_{S} = \frac{17}{4}$

$\Leftrightarrow a = b = \frac{1}{2}$
Kaya Renger Coordinator 22/09/2017 at 16:39

No :< , we can do this !

Denote $t=\dfrac{1}{ab}$ ; $\dfrac{1}{t}=ab$

$t=\dfrac{1}{ab}\ge\dfrac{1}{\left(\dfrac{a+b}{2}\right)^2}=\dfrac{1}{\left(\dfrac{1}{2}\right)^2}=4$

Predict a = b = $\dfrac{1}{2}$ then t = 4 , we have :

$S=\dfrac{1}{t}+t=\left(\dfrac{1}{t}+\dfrac{t}{16}\right)+\dfrac{15t}{16}\ge2.\sqrt{\dfrac{1}{t}.\dfrac{t}{16}}+\dfrac{15.4}{16}=\dfrac{17}{4}$

So $Min_S=\dfrac{17}{4}$

$\Leftrightarrow a=b=\dfrac{1}{2}$
Lê Quốc Trần Anh Coordinator 22/09/2017 at 16:44

Oh, I was wrong! Thanks Dao Trong Luan and Kaya Renger :)
Dao Trong Luan 22/09/2017 at 16:43

a,b are positive number, aren't integer, so they can are positive fraction

So Lê Quốc Trần Anh is wrong
Lê Quốc Trần Anh Coordinator 22/09/2017 at 15:30

Because a,b are positive numbers and a + b = 1

=> a = b = 0 or a = 1; b = 0 or a = 0;b = 1.

If a = b = 0 then $\dfrac{1}{ab}=\dfrac{1}{0}$ (unsatisfy)

If a or b = 0 or 1, then $\dfrac{1}{ab}=\dfrac{1}{0}$ (unsatisfy)

So there is no $a,b$ satisfy.

Kaya Renger Coordinator

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