<s>[INST] How to show that $\log (\frac{2a}{1-a^2}+\frac{2b}{1-b^2}+\frac{2c}{1-c^2})= \log\frac{2a}{1-a^2}+ \log \frac{2b}{1-b^2}+ \log \frac{2c}{1-c^2}$ If $\log (a +b +c) =\log a+\log b+\log c$ then show that $$\log \left(\frac{2a}{1-a^2}+\frac{2b}{1-b^2}+\frac{2c}{1-c^2}\right)= \log\frac{2a}{1-a^2}+ \log \frac{2b}{1-b^2}+ \log \frac{2c}{1-c^2}$$
Trial: If put $a=\frac{2a}{1-a^2}$ and the similar we are done. Can we do this in this way? Otherwise how to do. Please help. 
 [/INST] OK, here's another way to do it:
Let $a\circ b=\dfrac{a+b}{1-ab}$.
Prove that this operation is associative.
Show that $a\circ b\circ c = \dfrac{a+b+c-abc}{1-ab-ac-bc}$.
That implies that if $a+b+c=abc$, then $a\circ b\circ c=0$.
It's easy to show that $0\circ 0=0$, so we have
\begin{align}
0 & = 0\circ 0 = (a\circ b\circ c)\circ(a\circ b\circ c) \\[12pt]
& = (a\circ a)\circ(b\circ b)\circ(c\circ c) \qquad (\text{This is where associativity is used.}) \\[12pt]
& = \frac{2a}{1-a^2}\circ\frac{2b}{1-b^2} \circ\frac{2c}{1-c^2}.
\end{align}
So that implies the numerator $\dfrac{2a}{1-a^2}+\dfrac{2b}{1-b^2} +\dfrac{2c}{1-c^2} -\dfrac{2a}{1-a^2}\cdot\dfrac{2b}{1-b^2} \cdot\dfrac{2c}{1-c^2}$, is $0$.
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