Posted by shivani on Sunday, June 29, 2014 at 5:45am.
if abc=1 prove that 1/(1+a+b^1) + 1/(1+b+c^1) +1/(1+c+a^1) =1

maths (exponents  Steve, Sunday, June 29, 2014 at 4:32pm
Well, just using brute force, we have
b/(b+ab+1) + c/(c+bc+1) + a/(a+ac+1)
expand all that over a common denominator, and you get
(a^2b^2c + a^bc^2 + 2a^2bc + a^2b + ab^2c^2 + 2ab^2c + 2abc^2 + 6abc + 2ab + ac^2 + 2ac + a + b^2c + 2bc + b + c)
/
(a^2b^2c^2 + a^2b^2c + a^2bc^2 + 2a^2bc + a^2b + ab^2c^2 + 2ab^2c + 2abc^2 + 4abc + 2ab + ac^2 + 2ac + a + b^2c + 2bc + b + c + 1)
now replace all the abc occurrences with just 1, and you have
(ab + ac + 2a + a^2b + bc + 2b + 2c + 6 + 2ab + ac^2 + 2ac + a + b^2c + 2bc + b + c)
/
(ab + ac + 2a + a^2b + bc + 2b + 2c + 6 + 2ab + ac^2 + 2ac + a + b^2c + 2bc + b + c)
where the top and bottom are clearly the same.
Now, for a trick of evaluation which avoids all that mess, I haven't come up with one yet.
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