I have to show that $\forall n \in \mathbb{N}, n \ge 2$ if we have
$$x_1, x_2, ... x_n > 0$$
and
$$x_1 \cdot x_2 \cdot ... \cdot x_n = 1$$
then it is true that
$$x_1 + x_2 + ... x_n \ge n$$
How can I prove this? The textbook recommends using induction, but after trying it, I didn't get anywhere.
If you want to avoid A.M. - G.M. inequality you can prove it by induction as follows.
Suppose the result is true for $n$. Consider positive numbers $x_1,x_2,...,x_n,x_{n+1}$ whose product is $1$.
Let $y_i=x_i$ for $i <n$ and $y_n=x_nx_{n+1}$. Then $y_i$'s are positive and their product is $1$. By induction hypothesis we get $y_1+y_2+...+y_n\geq n$. Now $$x_1+x_2+...+x_n+x_{n+1}= y_1+y_2+..+y_n+x_{n+1} -x_nx_{n+1}$$ $$ \geq n +x_n+x_{n+1}-x_n+x_{n+1}$$ $$ \geq n+1$$ because $((1-t)(1-s) \leq 1$, so $t+s=ts \geq ts$ for $0,\leq t,s \leq 1$. [Take $t-x_n,s=x_{n+1}$].
Prof using AM -GM inequality: $\frac {x_1+x_2+...+x_n} n \geq (x_1x_2...x_n)^{1/n}=1^{1/n}=1$ so $x_1+x_2+...+x_n \geq 1$.
