Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Have a look at this:…. 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

Somehow I can't write a comment. That's why I write this as an answer - sorry...

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!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.