Prove in as elementary a way as possible that $a^{1/(a-1)} < 2$ for integer $a \ge 3$

Question

This inequality came up as the final step in an elementary proof that the only integer solution to $x^y = y^x$ with $x > y > 1$ is $4^2 = 2^4$.

Here are two proofs I came up with that if integer $a \ge 3$, then $ a^{1/(a-1)} < 2$. Both proofs do it by noting that this is true for $a=3$, and $a^{1/(a-1)}$ is decreasing for $a \ge 3$.

One way to show this is by looking at the log of this, which is $\frac{\ln a}{a-1}$. Its derivative is $\frac{(a-1)/a - \ln a}{(a-1)^2} =\frac{1-1/a - \ln a}{(a-1)^2} < 0$ for $a \gt 1$. If $f(a) =1-1/a - \ln a $ then $f(1) = 0$ and $f'(a) =1/a^2-1/a < 0 $ for $a > 1$.

Another proof: Suppose $a^{1/(a-1)} \le (a+1)^{1/a} $. Then $a^a \le (a+1)^{a-1}$ or $(a+1)a^a \le (a+1)^a$. Dividing by $a^a$, $a+1 \le (1+1/a)^a$. But, as has been proved here by elementary means many times, $(1+1/a)^a < 3$, so that $a+1 < 3$ or $a < 2$.

So, is there a more elementary proof than these?

Thanks.

Answer 1

The inequality $a^{1/(a-1)}<2$ for integers $a\ge 3$ is equivalent to $a<2^{a-1}$ for integers $a\ge 3$. This is easily proved by induction on $a$. It’s clearly true for $a=3$, and if $a<2^{a-1}$ and $a\ge 3$, then

$$a+1<2^{a-1}+1<2\cdot 2^{a-1}=2^a\;.$$

Answer 2

Inasmuch as Bernoulli's strict inequality counts as elementary:

$(1 +x)^r \gt 1 + r x$ for $r \in \mathbb{N}, r \ge 2$ and $x \in \mathbb{R}, x \ge -1, x \ne 0$

$a \lt 2^{a-1}$ for $a \ge 3$ follows directly from it with $r = a-1 \ge 2$ and $x=1 \gt 0$.