a^x = bc, b^y = ca, c^z = ab. Prove that, (x/(1 + x)) + (y/(1 + y)) + (z/(1 + z)) = 2. (Without using log) a ≠ b ≠ c


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Question Number 191589 by MATHEMATICSAM last updated on 28/Apr/23

$a^{x} = b c, b^{y} = c a, c^{z} = a b .$ $Prove that, \frac{x}{1 + x} + \frac{y}{1 + y} + \frac{z}{1 + z} = 2 .$ $(Without using \log)$ $a \neq b \neq c$
Commented by mr W last updated on 27/Apr/23

$n o t t r u e i f$ $a = \frac{1}{2}, b = 1, c = 2$ $x = y = z = - 1$
	Answered by Rasheed.Sindhi last updated on 26/Apr/23

	$a^{x} = b c, b^{y} = c a, c^{z} = a b .$ $Prove that, \frac{x}{1 + x} + \frac{y}{1 + y} + \frac{z}{1 + z} = 2 .$ $a^{x} b^{y} c^{z} = (b c) (c a) (a b) = a^{2} b^{2} c^{2}$ $x = 2, y = 2, z = 2$ $\frac{x}{1 + x} + \frac{y}{1 + y} + \frac{z}{1 + z}$ $= \frac{2}{1 + 2} + \frac{2}{1 + 2} + \frac{2}{1 + 2}$ $= \frac{2}{3} + \frac{2}{3} + \frac{2}{3} = \frac{2}{3} \times 3 = 2$
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