induction exercise doubt

Question

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.

Answer 1

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!

Answer 2

For the sake of completeness I post an answer:

Without induction:

$$ x_1 x_2 \dots x_n = 1 \iff (x_1 x_2 \dots x_n)^{1/n} = 1$$

then by the AM-GM inequality (that can be proven using induction) we have that

$$ \frac{x_1 +x_2 + \dots + x_n}{n} \ge (x_1 x_2 \dots x_n)^{1/n} = 1$$ and we are done.

With induction:

If $x_1 x_2 \dots x_n = 1$ there are two cases, either $x_k\ge 1, x_m \ge1$ for some $k,m \in \{1,2, \dots, n \}$ or $x_k\le 1, x_m \le1$ for some $k,m \in \{1,2, \dots, n \}$.

we assume the inductive step, i.e., that if $x_1 x_2 \dots x_{n-1} =1$ then $\sum_{i=1}^{n-1} x_i > n - 1$.

Now assume we are in the case $x_k \ge 1, x_m \ge 1$ (there other case is similar) put $x'_k = x_k x_m$ then from the inductive step we have that

$$ \sum_{i=1}^{n-2} x_i + x'_k \ge n-1 $$ and noticing that $$ x_kx_m + 1 - x_k -x_m = (x_k -1) (x_m -1) \ge 0$$ we obtain $$ \sum_{i=1}^{n} x_i -1 \ge \sum_{i=1}^{n-2} x_i + x'_k \ge n-1 $$