If log(a + b + c) = loga + logb + logc then prove that log(((2a)/(1 − a^2 )) + ((2b)/(1 − b^2 )) + ((2c)/(1 − c^2 ))) = log(((2a)/(1 − a^2 ))) + log(((2b)/(1 − b^2 ))) + log(((2c)/(1 − c^2 ))).


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Question Number 206332 by MATHEMATICSAM last updated on 12/Apr/24

$If \log (a + b + c) = \log a + \log b + \log c$ $then prove that$ $\log (\frac{2 a}{1 - a^{2}} + \frac{2 b}{1 - b^{2}} + \frac{2 c}{1 - c^{2}}) =$ $\log (\frac{2 a}{1 - a^{2}}) + \log (\frac{2 b}{1 - b^{2}}) + \log (\frac{2 c}{1 - c^{2}}) .$
	Answered by TheHoneyCat last updated on 12/Apr/24

	$Exponantiating we get the equivalent$ $problem :$ $S h o w t h a t (a + b + c) = a b c$ $\Rightarrow \frac{2 a}{1 - a^{2}} + \frac{2 b}{1 - b^{2}} + \frac{2 c}{1 - c^{2}}$ $= \frac{2 a}{1 - a^{2}} \times \frac{2 b}{1 - b^{2}} \times \frac{2 c}{1 - c^{2}}$ $\Leftrightarrow \frac{2}{1 - a^{2}} (a + \frac{b - a^{2} b}{1 - b^{2}} + \frac{c - a^{2} c}{1 - c^{2}})$ $= \frac{2 a}{1 - a^{2}} \times \frac{2 b}{1 - b^{2}} \times \frac{2 c}{1 - c^{2}}$ $\Leftrightarrow a + \frac{b - a^{2} b}{1 - b^{2}} + \frac{c - a^{2} c}{1 - c^{2}} = a \frac{2 b}{1 - b^{2}} \times \frac{2 c}{1 - c^{2}} \lor a^{2} = 1$ $\Leftrightarrow a - {a b}^{2} + b - a^{2} b + \frac{c - a^{2} c - b^{2} c + a^{2} b^{2} c}{1 - c^{2}}$ $= 2 a b \times \frac{2 c}{1 - c^{2}} \lor a = 1 \lor b = 1$ $\Leftrightarrow a - {a b}^{2} - {a c}^{2} + {a b}^{2} c^{2} + b - a^{2} b - {b c}^{2} + a^{2} {b c}^{2}$ $+ c - a^{2} c - b^{2} c + a^{2} b^{2} c = 2^{2} a b c$ $\lor a = 1 \lor b = 1 \lor c = 1$ $\Leftrightarrow a + b + c - ({a b}^{2} + {a c}^{2} + a^{2} b + {b c}^{2} + a^{2} c + b^{2} c) +$ $a b c (b c + a c + a b) = 2^{2} a b c \lor a = 1 \lor b = 1 \lor c = 1$ $\Leftrightarrow {a b}^{2} + {a c}^{2} + a^{2} b + {b c}^{2} + a^{2} c + b^{2} c$ $= a b c (a b + a c + b c - 3) \lor a = 1 \lor b = 1 \lor c = 1$ $\Leftrightarrow {a b}^{2} + {a c}^{2} + a^{2} b + {b c}^{2} + a^{2} c + b^{2} c$ $= a^{2} b + a^{2} c + a b c + {a b}^{2} + a b c + b^{2} c + a b c + {a c}^{2} + {b c}^{2}$ $- 3 a b c \lor a = 1 \lor b = 1 \lor c = 1$ $\Leftrightarrow {a b}^{2} + {a c}^{2} + a^{2} b + {b c}^{2} + a^{2} c + b^{2} c$ $= a^{2} b + a^{2} c + {a b}^{2} + b^{2} c + {a c}^{2} + {b c}^{2}$ $\lor a = 1 \lor b = 1 \lor c = 1$ $\Leftrightarrow True \lor a = 1 \lor b = 1 \lor c = 1$ $\Leftrightarrow True$ $So reading things backward . . .$ $If \log (a + b + c) = \log a + \log b + \log c$ $\log (\frac{2 a}{1 - a^{2}} + \frac{2 b}{1 - b^{2}} + \frac{2 c}{1 - c^{2}}) =$ $\log (\frac{2 a}{1 - a^{2}}) + \log (\frac{2 b}{1 - b^{2}}) + \log (\frac{2 c}{1 - c^{2}})_{◻}$
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