Question 1210195
<pre>

If a,b,c are in Harmonic Progression,

That means that 1/a, 1/b, 1/c are in arithmetic progression, i.e.

{{{1/b-1/a=1/c-1/b}}}

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

We are to show that

(1)   {{{1^""/(1/b + 1(c+a))- 1^""/(1/a + 1/(b+c))}}}{{{""=""}}}{{{1^""/(1/c + 1(a+b) )-1^""/(1/b + 1(c+a))}}}


All terms are of the form {{{1^""/(1/x + 1/(y+z))}}}

Multiply top and bottom by x(y+z)

{{{(x(y+z))/(y+z+x))}}}

So (1) becomes

(2)   {{{(b(c+a))/(c+a+b)-(a(b+c))/(b+c+a)}}}{{{""=""}}}{{{(c(a+b))/(a+b+c)-(b(c+a))/(c+a+b)}}} 

(2) will be true if and only if

{{{(2b(c+a))/(c+a+b)-(a(b+c))/(b+c+a)}}}{{{""=""}}}{{{(c(a+b))/(a+b+c)}}} 

which will be true if and only if

{{{2b(c+a)-a(b+c)}}}{{{""=""}}}{{{c(a+b)}}}

which will be true if and only if 

{{{2bc+2ab-ab-ac}}}{{{""=""}}}{{{ac+bc}}}

which will be true if and only if

{{{bc+ab-2ac}}}{{{""=""}}}{{{0}}}

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

{{{((2ac)/(a+c))c+a((2ac)/(a+c))-2ac}}}{{{""=""}}}{{{0}}}

which will be true if and only if

{{{2ac^2+2a^2c-2ac(a+c) = 0}}}

Which will be true if and only if 

{{{c+a - (a + c) = 0}}}

And since this is true, the problem is proved.

Edwin</pre>