What′s the smallest value of a^2 +b^2 +(1/(ab)) for a, b>0?


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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

	$a n o t h e r w a y$ $f (a, b) = a^{2} + b^{2} + \frac{1}{a b}$ $\frac{\partial f}{\partial a} = 2 a - \frac{1}{a^{2} b} = 0 \Rightarrow 2 a^{3} b = 1 . . . (i)$ $\frac{\partial f}{\partial b} = 2 b - \frac{1}{{a b}^{2}} = 0 \Rightarrow 2 {a b}^{3} = 1 . . . (i i)$ $f r o m (i) a n d (i i) :$ $a = b = \frac{1}{\sqrt[4]{2}}$ $\Rightarrow f_{m i n} = 2 \sqrt{2}$ $({i t}^{'} s m i n i m u m, n o t m a x i m u m$ $b e c a u s e \frac{\partial^{2} f}{\partial a^{2}} > 0 a n d \frac{\partial^{2} f}{\partial b^{2}} > 0)$
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