Question 1043663
We wish to show that {{{log((a+b+c))=loga+logb+logc}}} implies 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)))}}}.

For the result to have meaning, it must be that 0 < a,b,c < 1, which can be determined by solving the inequalities
{{{2a/(1-a^2)>0}}}, {{{2b/(1-b^2)>0}}}, and {{{2c/(1-c^2)>0}}}.

With this in mind, it is enough to show that 

a+b+c = abc  ===> {{{a/(1-a^2) + b/(1-b^2) + c/(1-c^2) = 
(4abc)/((1-a^2)(1-b^2)(1-c^2))}}}.  <---- Why?


Now on to the proof:

{{{-(ab+ac+bc)abc + (ab+bc+ac)abc = 0}}}

===>{{{-(ab+ac+bc)(a+b+c) + (ab+bc+ac)abc = 0}}}, since a+b+c = abc,

===> {{{-ab(a+b+c) - ac(a+b+c)- bc(a+b+c) + (ab+bc+ac)abc = 0}}}

===> {{{-ab(a+b+c) + abc - ac(a+b+c) + abc - bc(a+b+c) + abc +  (ab+bc+ac)abc= 3abc}}}

===> {{{-ab(a+b) -ac(a+c)-bc(b+c) + (ab+bc+ac)abc = 3abc}}}

<===> {{{-a^2b - ab^2 - a^2c - ac^2 - b^2c-bc^2 + (ab+bc+ac)abc = 3abc}}}

<===>  {{{- ab^2- ac^2 - a^2b - bc^2  - a^2c - b^2c  + (ab+bc+ac)abc = 3abc}}} after a little rearrangement of terms.

Since a+b+c = abc, we get 

 {{{(a- ab^2- ac^2)+ (b - a^2b - bc^2) +(c - a^2c - b^2c)  + (ab+bc+ac)abc = 4abc}}}

===> {{{(a- ab^2- ac^2 + ab^2c^2)+ (b - a^2b - bc^2 + ba^2c^2) +(c - a^2c - b^2c + ca^2b^2) = 4abc}}}

<===> {{{a(1- b^2- c^2 + b^2c^2)+ b(1 - a^2 - c^2 + a^2c^2) + c(1 - a^2 - b^2 + a^2b^2) = 4abc}}}

<===> {{{a(1-b^2)(1-c^2) + b(1-a^2)(1-c^2) + c(1-a^2)(1-b^2) = 4abc}}}

<===>  {{{a/(1-a^2) + b/(1-b^2) + c/(1-c^2) = (4abc)/((1-a^2)(1-b^2)(1-c^2))}}}, after dividing both sides by {{{ (1-a^2)(1-b^2)(1-c^2)}}}


And that's it...