If a+b+c =1 then prove that (1/a - 1)(1/b - 1)(1/c - 1) = 0..........
A anjann New member Joined Aug 8, 2013 Messages 1 Aug 8, 2013 #1 If a+b+c =1 then prove that (1/a - 1)(1/b - 1)(1/c - 1) = 0..........
D Deleted member 4993 Guest Aug 8, 2013 #2 anjann said: If a+b+c =1 then prove that (1/a - 1)(1/b - 1)(1/c - 1) = 0.......... Click to expand... If a + b + c = 0 then prove that \(\displaystyle \displaystyle \left (\frac{1}{a} \ - \ 1 \right ) \ * \ \left(\frac{1}{b} \ - \ 1\right ) \ * \ \left(\frac{1}{c} \ - \ 1\right ) \ = 0\) In general, the above conjecture is not true. counterexample: let a = b = c = 1/3 → 1/a = 1/b = 1/c = 3 → \(\displaystyle \displaystyle \left (\frac{1}{a} \ - \ 1 \right ) \ * \ \left(\frac{1}{b} \ - \ 1\right ) \ * \ \left(\frac{1}{c} \ - \ 1\right ) \ = 8\) Last edited by a moderator: Aug 8, 2013
anjann said: If a+b+c =1 then prove that (1/a - 1)(1/b - 1)(1/c - 1) = 0.......... Click to expand... If a + b + c = 0 then prove that \(\displaystyle \displaystyle \left (\frac{1}{a} \ - \ 1 \right ) \ * \ \left(\frac{1}{b} \ - \ 1\right ) \ * \ \left(\frac{1}{c} \ - \ 1\right ) \ = 0\) In general, the above conjecture is not true. counterexample: let a = b = c = 1/3 → 1/a = 1/b = 1/c = 3 → \(\displaystyle \displaystyle \left (\frac{1}{a} \ - \ 1 \right ) \ * \ \left(\frac{1}{b} \ - \ 1\right ) \ * \ \left(\frac{1}{c} \ - \ 1\right ) \ = 8\)
