if xyz=1, prove ((x/(x−1)))^2 +((y/(y−1)))^2 +((z/(z−1)))^2 ≥1.


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Question Number 196629 by universe last updated on 28/Aug/23

$i f x y z = 1, p r o v e$ ${(\frac{x}{x - 1})}^{2} + {(\frac{y}{y - 1})}^{2} + {(\frac{z}{z - 1})}^{2} ⩾ 1 .$
Commented by universe last updated on 28/Aug/23

$You can't use 'macro parameter character #' in math mode$
	Answered by universe last updated on 28/Aug/23

	$l e t a = \frac{x}{1 - x}, b = \frac{y}{1 - y}, c = \frac{z}{1 - z}$ $a b c = \frac{x y z}{(1 - x) (1 - y) (1 - z)} = \frac{1}{(1 - x) 1 - y) (1 - z)}$ $a + 1 = \frac{1}{1 - x}, b + 1 = \frac{1}{1 - y}, c + 1 = \frac{1}{1 - z}$ $(a + 1) (b + 1) (c + 1) = \frac{1}{(1 - x) (1 - y) (1 - z)}$ $(a + 1) (b + 1) (c + 1) = a b c$ $a b c = a b c + a b + b c + c a + a + b + c + 1$ $a b + b c + c a + a + b + c + 1 = 0$ $2 (a b + b c + c a) + 2 (a + b + c) + 2 = 0$ $a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a) + 2 (a + b + c) + 2 = a^{2} + b^{2} + c^{2}$ ${(a + b + c)}^{2} + 2 (a + b + c) + 2 = a^{2} + b^{2} + c^{2}$ ${(a + b + c + 1)}^{2} + 1 = a^{2} + b^{2} + c^{2}$ $n o w {(a + b + c + 1)}^{2} + 1 ⩾ 1$ $s o a^{2} + b^{2} + c^{2} ⩾ 1$ ${(\frac{x}{1 - x})}^{2} + {(\frac{y}{1 - y})}^{2} + {(\frac{z}{1 - z})}^{2} ⩾ 1$
	Commented by mr W last updated on 28/Aug/23

	$g r e a t! t h a n k s!$
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