```Question 522562
<pre>
a+b+c = {{{1/(a+b+c)}}}

Multiply both sides by (a+b+c)

(a+b+c)² = 1

Take positive square roots of both sides:

a+b+c = 1

So we just need three fractions that have sum = 1

To do that take any 3 integers, say 1, 4 and 5

Put their sum over itself and that will equal 1

{{{(1+4+5)/(1+4+5)}}} = 1

Add the terms in the denominator but leave the numerator

{{{(1+4+5)/10}}} = 1

Write as the sum of three fractions:

{{{1/10}}} + {{{4/10}}} + {{{5/10}}} = 1

Reduce any that will reduce:

{{{1/10}}} + {{{2/5}}} + {{{1/2}}} = 1

So  a={{{1/10}}}, b={{{2/5}}}, c={{{1/2}}}

------------------

There is no end to the number of different possibilities following that recipe:

{{{(2+3+4)/(2+3+4)}}} = {{{(2+3+4)/9}}} = {{{2/9}}} + {{{3/9}}} + {{{4/9}}} = {{{2/9}}} + {{{1/3}}} + {{{4/9}}} = 1

{{{(1+2+3)/(1+2+3)}}} = {{{(1+2+3)/6}}} = {{{1/6}}} + {{{2/6}}} + {{{3/6}}} = {{{1/6}}} + {{{1/3}}} + {{{1/2}}} = 1

{{{(1+2+5)/(1+2+5)}}} = {{{(1+2+5)/8}}} = {{{1/8}}} + {{{2/8}}} + {{{5/8}}} = {{{1/8}}} + {{{1/4}}} + {{{5/8}}} = 1

Edwin</pre>```