If $a \ge 1$ then {$a^{1/n}$} is a decreasing real sequence

Question

In the proof of theorem 16.4 in Foundations of Mathematical Analysis by Johnsonbaugh and Pfaffenberger, the authors state without proof that if $a \ge 1$ then {$a^{1/n}$} is a decreasing real sequence. As an exercise I decided to prove this by induction. But I'm stuck trying to prove that $S(n) \Rightarrow S(n+1)$. Question: how does one prove the inductive step?

$n \in \mathbb{P}$

$a,r,q,z,y,w \in \mathbb{R}$

$S(n)$ is $(z^n = y^{n+1} = a \ge 1) \Rightarrow (z \ge y)$

Suppose $w^{n+2} = a$,

$S(n+1)$ is $(y^{n+1} = w^{n+2} = a \ge 1) \Rightarrow (y \ge w)$

Suppose $r^1 = q^2 = a$,

$S(1)$ is $(r = q^2 \ge 1) \Rightarrow (r \ge q)$

$0 < q < 1 \Rightarrow q^2 < 1$, therefore $q \ge 1$ and $r = q^2 \ge q$

So the base case is true...

Answer 1

You don't really need induction. Using that $a=a^1=a^{1/(n+1)+n/(n+1)}$, $$ a^{1/(n+1)}\leq a^{1/n}\ \iff\ a^{n/(n+1)}\leq a\ \iff\ a^{n/(n+1)}\leq a^{1/(n+1)}\,a^{n/(n+1)}\ \iff\ 1\leq a^{1/(n+1)}\ \iff\ 1\leq a. $$ The last if and only if is the non obvious one. But just notice, for $\implies$, that if $a^{1/(n+1)}\geq1$ then so is the product with itself $n$ times. And if it is less than $1$, then so is the product with itself $n$ times, which proves the converse.

Answer 2

By contradiction. Let $$x^{n+1} = y^n = a$$ $$0 \le z \lt 1 \Rightarrow z^m \lt 1 \Rightarrow a^{\frac1m} \ge 1$$ $$x \gt y \ge 1 \Rightarrow x^{n+1} \gt x^n \gt y^n \Rightarrow a^{\frac{1}{n+1}} \le a^{\frac1n}$$