if x;y;z≥1 then: (1/(3xy−1)) + (1/(3yz−1)) + (1/(3zx−1)) ≥ (3/(2xyz))


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Question Number 147488 by mathdanisur last updated on 21/Jul/21

$i f x; y; z ⩾ 1 t h e n :$ $\frac{1}{3 x y - 1} + \frac{1}{3 y z - 1} + \frac{1}{3 z x - 1} ⩾ \frac{3}{2 x y z}$
	Answered by mindispower last updated on 21/Jul/21

	$i n f i r t$ $1 ⩽ x y, x y = \frac{x y z}{z}$ $\Rightarrow \frac{1}{3 x y - 1} ⩾ \frac{1}{\frac{3 x y z}{z} - x y} = \frac{z}{2 x y z}$ $s a m i d e a \Rightarrow \frac{1}{3 y z - 1} ⩾ \frac{x}{2 x y z}, \frac{1}{3 x z - 1} ⩾ \frac{y}{2 x y z}$ $L H ⩾ \frac{1}{2 x y z} (x + y + z)$ $s i n c x : y : z ⩾ 1$ $\Rightarrow x + y + z ⩾ 3$ $\Rightarrow \frac{1}{3 x y - 1} + \frac{1}{3 z x - 1} + \frac{1}{3 y z - 1} ⩾ \frac{1.3}{2 x y z} = \frac{3}{2 x y z}$
	Commented by mathdanisur last updated on 21/Jul/21

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