induction exercise doubt

the exercise states:

Let $x_1 , ...,x_n$ be strictly positive numbers such that their product is equal to 1. Show then that $\sum_{k=1}^{n} {x_k} \ge n$, for every $n \ge 2$.

My solution:

for the base case split in two cases, $x_1 = x_2 = 1$ and $0<x_1<1<x_2$

for the first case $x_1 + x_2 = 1+1 \ge 2$ for the second case $x_1 + x_2 \ge 2 \implies x_1+x_2 -x_1x_2 -1 \ge 2 -x_1x_2 -1 \implies (1-x_1) (x_2 -1) \ge 0$ wich is true.

So the base cases are gone (the base case $0<x_2<1<x_1$ is true by symmetry).

Now we assume $\sum_{k=1}^{n} {x_k} \ge n$ is true and try to prove $\sum_{k=1}^{n+1} {x_k} \ge n+1$.

$\sum_{k=1}^{n+1} {x_k} \ge n+1 \iff \sum_{k=1}^{n} {x_k} + x_{n+1} \ge n+1 \iff \sum_{k=1}^{n} {x_k} \ge n+1 -x_{n+1}$

(here is where I am unsure and I get a little wordy) Now the only value that the$x_{n+1}$ can have is 1 because the product of $x_1x_2...x_n$ = 1 so if we add a single number to this product and we want to keep the product constant the only value $x_{n+1}$ can have is 1.

So $\sum_{k=1}^{n} {x_k} \ge n+1 -x_{n+1} \iff \sum_{k=1}^{n} {x_k} \ge n$

And we are done.

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Have a look at this: en.wikipedia.org/wiki/…. One of those proofs should help you out. – Prahlad Vaidyanathan Aug 11 '14 at 13:21
Hint: When do induction, you can not change the assumption, in your case, the assumption, $x_1x_2... x_{n+1}=1$ is the assumption, but you could translate your assumption into different form. Try set $x_i'=x_i for 1\leq i\leq n-1, x_n'=x_nx_{n+1}$. – ougao Aug 11 '14 at 13:24

The comments given above by the other guys should shove your problem anyways. I just wanted to add, that in the induction step you can't assume that $x_{n+1}=1$, since you only know that $\prod_{i=1}^{n+1}x_i=1$. You don't know what the product of the first n factors is!