proof by induction: product of number = 1

Question

Before anything, sorry if I make any phrasing mistakes. I'm not 100% comfortable with writing in English.

So here's my problem. I don't want the resolution but just a hint on how to proceed for solving it.

For any $n \in \mathbb N$, for any $r_1, r_2, \dots , r_n \in \mathbb R$ such that $\forall i, 1 \le i \le n, r_i > 0$ and $r_1 \times r_2 \times \dots \times r_n = 1$, if $\exists i, 1 \le i \le n$ such that $r_i < 1$, then $\exists j, 1 \le j \le n$ such that $r_j > 1$.

I'm pretty sure it's an induction problem since I need to prove that the hypothesis is true for any $n$.

How I started:

$A(n) = (\exists i, 1 \le i \le n ∣ r_i < 1) \rightarrow (\exists j, 1 \le j \le n ∣ r_j > 1)$

Hypothesis: For any $n \in \mathbb N$, for any $r_1, r_2, \dots , r_n \in \mathbb R$ such that $\forall i, 1 \le i \le n, r_i > 0$ and $r_1 \times r_2 \times \dots \times r_n = 1, A(n)$

Basis step ($ n=1 $):

We have $r_1 = 1$ since the product of the $r_i $ equals $ 1$, therefore since there's only one number ($r_1 = 1$) there's no number $ \ne 1$. Therefore we have $\mathrm {fasle} \rightarrow \mathrm {false}$, so $A(1)$ is true. The basis step is verified.

Let's suppose $A(n)$ is true for $n = k$ (an arbitrary positive integer), then we must show that $A(n)$ is true for $n = k + 1$.

And this is where I'm blocked. I can't find a way to show that if there's a number $< 1$ in the $r_i$ there's necessarily a number $> 1$. And maybe I've made a mistake in my basis step but I can't find where.

Any hint on how to progress or why my reasoning is wrong would be very appreciated.

Thank you very much!

Answer 1

It's actually easier: if $\exists i :r_i<1$ and if we assume that $0<r_j\leq1 \forall j\neq i$, then the product would be $<1\cdot1\cdot\dots1=1$

Answer 2

Try proving the equivalent contrapositive:

If $0 <r_i \le 1$ for all $i$ then $\prod r_i \le 1$ and if any $r_k<1$ then $\prod r_i <1$.

And that is easier if we prove the stronger:

If $0 <r_i\le 1$ for all $i $ then $\prod r_i \le \min (r_i)\le 1$.

Proof: $n=1$ obvious.

$r_1=\min(r_i)\le 1$

$n=2$. If $0 <r_1\le 1$ and $0 <r_2\le 1$, then $r_1r_2\le r_1*1=r_1\le 1$. Also $r_1r_2\le 1*r_2 =r_2\le 1$ so $r_1r_2\le \min (r_i)\le 1$

Inductive step:

If $\prod r_i \le \min r_i $ for $1\le i\le n $ and if $0 <r_{n+1}\le 1$ then

$(\prod r_i) *r_{n+1}\le 1*r_{n+1}=r_{n+1}\le 1$.

And $(\prod r_i) *r_{n+1}\le (\prod r_i*1\le \min (r_i)\le 1$.

So $(\prod r_i) *r_{n+1}\le \min (\min r_i, r_{n+1})=\min_{i\le n+1}r_i\le 1$.

QED.

This proves your result. If $\prod r_i =1$ but some $r_k <1$ then if all $r_i\le 1$ we'd have $\prod r_i \le r_k <1$. Since that didn't happen so $p_m > 1$ for some $m $.