If $ x_{1} := 1 $ and $ x_{n + 1} := x_{n} + \dfrac{n}{(x_{1} \times \cdots \times x_{n})^{1/n}} $, then $ \dfrac{x_{n}}{\ln(n)} \to \infty $. Define a sequence $ (x_{n})_{n \in \mathbb{N}} $ of positive real numbers by
$$
x_{1}     := 1 \quad \text{and} \quad
\forall n \in \mathbb{N}: \quad
x_{n + 1} := x_{n} + \frac{n}{(x_{1} \times \cdots \times x_{n})^{1/n}}.
$$
Define an associated sequence $ (y_{n})_{n \in \mathbb{N}_{\geq 2}} $ of positive real numbers by
$$
\forall n \in \mathbb{N}_{\geq 2}: \quad
y_{n} := \frac{x_{n}}{\ln(n)}.
$$
I believe that the Cauchy-Schwarz Inequality yields $ \displaystyle \lim_{n \to \infty} y_{n} = \infty $. Is my assertion correct? A proof of it may be found in another post, but I haven’t gotten down to checking it yet. Thanks for your help!

So
$x_{n+1} = x_n + \frac{n}{(\prod_{k=1}^n x_k)^{1/n}}$
and
$x_1 = 1$.
You claim that
$\frac{x_n}{\ln n}
\to \infty$
and would like a proof.
btw, what is the "CS theorem"?
Obviously the $x_n$ are increasing.
Therefore,
$x_1 < (\prod_{k=1}^n x_k)^{1/n} < x_n$
so that
$x_{n+1} 
> x_n + \frac{n}{x_n}
$
and
$x_{n+1} 
< x_n + n
$.
From the second inequality,
$x_n
< \frac{n(n+1)}{2}
$.
Putting this in the first inequality,
$x_{n+1} 
> x_n + \frac{n}{\frac{n(n+1)}{2}}
= x_n + \frac{2}{n+1}
$.
From this,
$x_n > 2\ln n+c$
for some real $c$.
This isn't enough,
but it's a start.
Suppose that $ x_n \sim  r n$.
Then,
$$\prod_{k=1}^n x_k
\sim r^n n!
\sim r^n \sqrt{2\pi n}\frac{n^n}{e^n}
= (nr/e)^n \sqrt{2\pi n}
$$
so
$$(\prod_{k=1}^n x_k)^{1/n}
\sim (nr/e) (2\pi n)^{1/(2n)}
$$
so
$$x_{n+1}
\sim x_n + \frac{n}{(nr/e) (2\pi n)^{1/(2n)}}
= x_n + \frac{e}{r (2\pi n)^{1/(2n)}}
.$$
Since
$(2\pi n)^{1/(2n)}
\to 1$,
if
$r =\frac{e}{r}$
(or $r = e^{1/2}$),
$x_{n+1}
\sim x_n + r$.
I think this shows that
$x_n \sim e^{1/2} n$,
which certainly implies
$\frac{x_n}{\ln n} \to \infty$.
I'll check this computationally tomorrow.
Now, time for sleep.
