Question 1072286
{{{a^x}}}{{{""=""}}}{{{bc}}};{{{b^y}}}{{{""=""}}}{{{ca}}};{{{c^z}}}{{{""=""}}}{{{ab}}} then prove {{{x/(x+1) + y/(y+1) + z/(z+1)}}}{{{""=""}}}{{{2}}}
<pre><b>
To avoid doing essentially the same thing three times, 
I'll just do it once with the letters p,q,r,s and use it as
a formula for each of the fractions:

{{{p^q}}}{{{""=""}}}{{{r*s}}}
{{{log((p^q))}}}{{{""=""}}}{{{log((r*s))}}}
{{{q*log((p))}}}{{{""=""}}}{{{log((r))+log((s))}}}
{{{q}}}{{{""=""}}}{{{(log((r))+log((s)))/(log((p)))}}}

Therefore,

{{{q/(q+1)=((log((r))+log((s)))/(log((p))))/(((log((r))+log((s)))/(log((p))))+1)}}}{{{""=""}}}{{{(log((r))+log((s)))/(log((r))+log((s))+log((p)))}}}

and therefore, using the above as a formula,

{{{x/(x+1) + y/(y+1) + z/(z+1)}}}{{{""=""}}}{{{(log((b))+log((c)))/(log((b))+log((c))+log((a))) + (log((c))+log((a)))/(log((c))+log((a))+log((b))) + (log((a))+log((b)))/(log((a))+log((b))+log((c)))}}}

{{{(2*log((a))+2*log((b))+2*log((c)))/(log((a))+log((b))+log((c)))}}}{{{""=""}}}{{{(2*(log((a))+log((b))+log((c))))/(log((a))+log((b))+log((c)))}}}{{{""=""}}}{{{(2*(cross(log((a))+log((b))+log((c)))))/(cross(log((a))+log((b))+log((c))))}}}{{{""=""}}}{{{2}}}

Edwin</pre></b>