Induction Method?

Question

Prove that for all n>1, if x[1], x[2],..., x[n] are real numbers strictly between 0 and 1, then (1-x[1])(1-x[2]0...91-x[n]> 1-x[1]-x[2]-...-x[n]

Answer 1

there is nothing to prove n >1 so there you go!

Answer 2

I think you had some typos, but assuming that 0 should be a ), the 9 should be a (, and there should be a ) after the -x[n], then
it is a true statement, but it should look like this:

(1-x[1]) * (1-x[2]) * ... * (1-x[n]) > 1-x[1] - x[2] - ... -x[n]

for any n >1, where x[n] is a real number between 0 and 1, the terms to the left of the > will approach 0 while the term to the right of it will continue to decrease at an increasing rate (becoming more negative).

As far as a proof, sorry, I wouldn't know where to start other than demonstrating/graphing the pattern and extrapolating from a sample set.

Answer 3

Hint:

Is it true for n=1?

Assuming it is true for n=k, can you show it for n=k+1? Just multiply.