if x;y;z>0 and (1/(1+x)) + (1/(1+y)) + (1/(1+z)) = 1 then prove that: x + y + z ≥ (3/4) xyz


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Question Number 161280 by HongKing last updated on 15/Dec/21

$if x; y; z > 0 and \frac{1}{1 + x} + \frac{1}{1 + y} + \frac{1}{1 + z} = 1$ $then prove that :$ $x + y + z ⩾ \frac{3}{4} xyz$
	Answered by 1549442205PVT last updated on 16/Dec/21

	$F r o m t h e h y p o t h e s i s \frac{1}{1 + x} + \frac{1}{1 + y} + \frac{1}{1 + z} = 1$ $w e g e t x y + y z + z x + 2 (x + y + z) + 3 = x y z + x y + y z + z x + x + y + z + 1$ $o r x + y + z + 2 = x y z . H e n c e, t h e i n e q u a l i t y$ $w h i c h w e n e e d p r o v e t o b e e q u i v a l e n t t o$ $4 (x + y + z) ⩾ 3 (x + y + z + 2) \Leftrightarrow x + y + z ⩾ 6 ()$ $F r o m t h e h y p o t h e s i s x + y + z + 2 = x y z,$ $a p p l y i n g t h e i n e q u a l i t y A M - G M w e$ $h a v e x + y + z + 2 = x y z ⩽ {(\frac{x + y + z}{3})}^{3}$ $\Rightarrow a^{3} - 27 a - 54 ⩾ 0 \Rightarrow (a - 6) (a^{2} + 6 a + 9) ⩾ 0$ $\Leftrightarrow (a - 6) {(a + 3)}^{2} ⩾ 0 \Rightarrow a = x + y + z ⩾ 6$ $T h u s, t h e i n e q u a l i t y () p r o v e d, s o$ $x + y + z ⩾ \frac{3}{4} x y z (q . e . d)$ $T h e e q u a l i t y o c c u r s i f a n d o n l y i f$ $x = y = z = 2$
	Commented by HongKing last updated on 17/Dec/21

	$perfect my dear Sir thank you$
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