Let $x_1,x_2,\ldots,x_n$ positive numbers such that $x_1*x_2*\ldots*x_n=1\\$
Prove by induction that:
$x_1+x_2+\ldots+x_n \ge n$
I have thought about dividing the lh side of the inequality by n, and then trying to use what was given to say that
$\frac{x_1+x_2+\ldots+x_n+x_{n+1}}{n+1} \ge\sqrt[n]{x_1*x_2\ldots*x_{n+1}} \ge 1 $
Am i heading the right direction? Any help would be appreciated!
Lemma: If the product of $n\geq 2$ positive reals is 1, then we can find one term that is $\leq 1$ and one term that is $\geq 1$.
Base case: Apply the lemma, so $(x_1 - 1)(x_2-1) \leq 0 \Rightarrow x_1 + x_2 \geq x_1 x_2 + 1 = 2$.
Induction step. Apply the lemma, WLOG let $x_n \leq 1, x_{n+1} \geq 1$.
Apply the induction hypothesis to the $n$ terms $x_1, x_2, \ldots, x_{n-1}, (x_nx_{n+1})$.
We have $ x_nx_{n+1} + \sum_{i=1}^{n-1} x_i \geq n$.
It remains to show that $x_n +x_{n+1} - x_n x_{n+1} \geq 1$, which is $0 \geq (x_n -1 ) ( x_{n+1} -1 ) $, which is obviously true by our WLOG definition.
(On second thought, I can't push through the forward part, so... let me think about it ...)
Hint: The "hard" part is proving the base case, namely $x_1 \times x_2 = 1 \Rightarrow x_1 + x_2 \geq 2$.
Hint: For these "inequalities similar to the AM-GM style", Forward Backward Induction works wonders.