Question 1083841
<pre>
We find the arithmetic mean of {{{1/a}}} and {{{1/c}}}

Add them {{{1/a+1/c=(a+c)/(ac)}}}

Divide by 2:  {{{(a+c)/(2ac)}}}

Its reciprocal {{{(2ac)/(a+c)}}} is the harmonic mean
between a and c, so

{{{b=(2ac)/(a+c)}}}

Substitute for b in

{{{1/(b-a) +1/(b-c)}}}

{{{1/(((2ac)/(a+c))-a) +1/(((2ac)/(a+c))-c)}}}

Multiply numerators and denominators by (a+c)

{{{(a+c)/(2ac-a(a+c)) +(a+c)/(2ac-c(a+c))}}}

{{{(a+c)/(2ac-a^2-ac) +(a+c)/(2ac-ac-c^2))}}}

{{{(a+c)/(ac-a^2) +(a+c)/(ac-c^2))}}}

{{{(a+c)/(a(c-a)) +(a+c)/(c(a-c)))}}}

{{{(a+c)/(a(c-a)) +(a+c)/(c(-c+a)))}}}

{{{(a+c)/(a(c-a)) +(a+c)/(-c(c-a)))}}}

{{{(a+c)/(a(c-a)) -(a+c)/(c(c-a)))}}}

Factor out {{{(a+c)/(c-a)}}}

{{{((a+c)/(c-a))(1/a-1/c)))}}}

{{{((a+c)/(c-a))((c-a)/(ac))))}}}

{{{((a+c)/(cross(c-a)))((cross(c-a))/(ac))))}}}

{{{((a+c)/(ac))))}}}

{{{a/(ac)+c/(ac)}}}

{{{cross(a)/(cross(a)c)+cross(c)/(a*cross(c))}}}

{{{1/c+1/a}}}

{{{1/a+1/c}}}

Edwin</pre>