Given a,b,c real and positive numbers, and a + b + c = 1 Find the minimum value of ((a + b)/(abc))


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Question Number 22154 by Joel577 last updated on 12/Oct/17

$Given a, b, c real and positive numbers, and$ $a + b + c = 1$ $Find the minimum value of \frac{a + b}{a b c}$
	Answered by ajfour last updated on 12/Oct/17

	$l e t a = c x a n d b = c y$ $t h e n c (x + y + 1) = 1$ $z = \frac{a + b}{a b c} = \frac{x + y}{c^{2} x y} = \frac{(x + y) {(x + y + 1)}^{2}}{x y}$ $A s z i s m i n i m u m,$ $\frac{\partial z}{\partial x} = \frac{x [{(x + y + 1)}^{2} + 2 (x + y) (x + y + 1)] - (x + y) {(x + y + 1)}^{2}}{x^{2} y} = 0$ $\Rightarrow x [x + y + 1 + 2 x + 2 y] - (x + y) (x + y + 1) = 0$ $3 x^{2} + 3 x y + x - x^{2} - y^{2} - 2 x y - x - y = 0$ $2 x^{2} + x y - y^{2} - y = 0$ $2 x^{2} + (x - 1) y - y^{2} = 0 . . . (i)$ $s i m i l a r l y \frac{\partial z}{\partial y} = 0 s h a l l g i v e$ $2 y^{2} + (y - 1) x - x^{2} = 0 . . . (i i)$ $\Rightarrow x = y$ $s o 2 x^{2} + x^{2} - x - x^{2} = 0$ $\Rightarrow 2 x^{2} - x = 0$ $o r x (x - \frac{1}{2}) = 0$ $w e s h a l l c h o o s e x = y = \frac{1}{2} s i n c e$ $a, b, a n d c a r e + v e r e a l n u m b e r s .$ $T h e n a = \frac{c}{2}; b = \frac{c}{2}$ $\Rightarrow \frac{c}{2} + \frac{c}{2} + c = 1$ $s o c = \frac{1}{2}, a = \frac{1}{4}, b = \frac{1}{4}$ $\Rightarrow m i n (\frac{a + b}{a b c}) = \frac{1 / 2}{1 / 32} = 16 .$
	Commented by Joel577 last updated on 12/Oct/17

	$T h a n k y o u v e r y m u c h S i r$
	Commented by Joel577 last updated on 12/Oct/17

	$I h a v e h e a r d a b o u t i n e q u a l i t y A M ⩾ G M$ $i s i t p o s s i b l e t o u s e t h a t i n e q u a l i t y ?$
	Answered by ajfour last updated on 12/Oct/17

	$l e t a + b = p$ $a n d f = \frac{a + b}{a b c} = \frac{p}{a b c} = \frac{4 p}{(4 a b) c}$ $\Rightarrow f = \frac{4 p}{[{(a + b)}^{2} - {(a - b)}^{2}] c}$ $o r f = \frac{4 p}{[p^{2} - {(a - b)}^{2}] c}$ $F o r f t o b e m i n i m u m, a = b$ $\Rightarrow f = \frac{4 p}{p^{2} c} = \frac{4}{p c}$ $b u t a + b + c = 1 \Rightarrow p + c = 1$ $m a x . o f c p r e s u l t s w h e n c = p$ $\Rightarrow c = \frac{1}{2} a n d p = \frac{1}{2}$ $S o f = \frac{4}{c p} = \frac{4}{(1 / 4)} = 16 .$
	Commented by Joel577 last updated on 13/Oct/17

	$t h a n k y o u v e r y m u c h$
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