What-s-the-smallest-value-of-a-2-b-2-1-ab-for-a-b-gt-0-

Question Number 180953 by depressiveshrek last updated on 19/Nov/22

{W h a t}^{'} s t h e s m a l l e s t v a l u e o f

a^{2} + b^{2} + \frac{1}{a b} f o r a, b > 0 ?

Answered by mr W last updated on 19/Nov/22

a^{2} + b^{2} + \frac{1}{a b} ⩾ 2 a b + \frac{1}{a b} ⩾ 2 \sqrt{2 a b \times \frac{1}{a b}} = 2 \sqrt{2}

m i n i m u m = 2 \sqrt{2}

w h e n a = b a n d 2 a b = \frac{1}{a b}, i . e . a = b = \frac{1}{\sqrt[4]{2}}

Answered by mr W last updated on 20/Nov/22

a n o t h e r w a y

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

\frac{\partial f}{\partial a} = 2 a - \frac{1}{a^{2} b} = 0 \Rightarrow 2 a^{3} b = 1 \dots (i)

\frac{\partial f}{\partial b} = 2 b - \frac{1}{{a b}^{2}} = 0 \Rightarrow 2 {a b}^{3} = 1 \dots (i i)

f r o m (i) a n d (i i) :

a = b = \frac{1}{\sqrt[4]{2}}

\Rightarrow f_{m i n} = 2 \sqrt{2}

({i t}^{'} s m i n i m u m, n o t m a x i m u m

b e c a u s e \frac{\partial^{2} f}{\partial a^{2}} > 0 a n d \frac{\partial^{2} f}{\partial b^{2}} > 0)

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