If $\frac{[x_n]^n [1-[x_n]^n]}{(1-x_n) n} = a$, is $[x_n]^n$ increasing for $n\geq 3$? Let $x_n$ be the solution to $\frac{x^n [1-x^n]}{(1-x) n} = a$, where $x \in [0,1], a \in [0,1]$ and $n \in \mathbf{N}$. I want to prove that $[x_n]^n$ is increasing in $n$ for $n\geq 3$.
(From numerical calculations I am quite confident that it should hold generally, but I would need a formal proof.)
 A: Choose an initial seed $x_0 \in [0, 1]$.
You want to show that
$$[x_{n+1}]^{n+1} \geq [x_n]^n,$$
for $n \geq 3$.
Since
$$\frac{{x_n}^n[1 - {x_n}^n]}{(1 - {x_n})n} = a$$
for $n \in \mathbb{N}$, we have
$$\frac{{x_{n+1}}^{n+1}[1 - {x_{n+1}}^{n+1}]}{(1 - {x_{n+1}})(n + 1)} = a.$$
Dividing through both equations in $a$:
$$\frac{{x_n}^n[1 - {x_n}^n]}{(1 - {x_n})n}\cdot\frac{(1 - {x_{n+1}})(n + 1)}{{x_{n+1}}^{n+1}[1 - {x_{n+1}}^{n+1}]} = 1.$$
Then the ratio $[x_{n+1}]^{n+1}/[x_n]^n$ takes the form:
$$\frac{[x_{n+1}]^{n+1}}{[x_n]^n} = \left(\frac{n + 1}{n}\right)\frac{1 + x_n + \ldots +  [x_n]^{n - 1}}{1 + x_{n+1} + \ldots + [x_{n+1}]^n},$$
where we've used $n \geq 3$ in the last equality.
Now, all you need to do is get a lower bound for
$$1 + x_n + \ldots [x_n]^{n - 1}$$
and an upper bound for
$$1 + x_{n+1} + \ldots + [x_{n+1}]^n.$$
You get, via the Arithmetic Mean-Geometric Mean Inequality:
$$1 + x_n + \ldots + [x_n]^{n - 1} > (n - 1)[x_n]^{(n/2)}$$
and
$$1 + x_{n+1} + \ldots + [x_{n+1}]^n \leq n,$$
since $x_n \in [0, 1]$.
We therefore have:
$$\frac{[x_{n+1}]^{n+1}}{[x_n]^n} > \left(\frac{n + 1}{n}\right)\left(\frac{(n - 1)[x_n]^{(n/2)}}{n}\right),$$
which simplifies to:
$$\frac{[x_{n+1}]^{n+1}}{[x_n]^n} > \left(1 + \frac{1}{n}\right)\left(1 - \frac{1}{n}\right)[x_n]^{(n/2)}.$$
I believe you can show this last quantity to be greater than or equal to $1$.
A: Thanks to Jose Arnaldo Dris, I found a solution:
First, since $x^n$ and $\frac{1-x^n}{n (1-x)} = \frac{1+\dots+x^{n-1}}{n}$ are both increasing in $x$ and decreasing in $n$, we have that $\frac{x^n [1-x^n]}{n (1-x)}$ is also increasing in $x$ and decreasing in $n$. Therefore $x_{n+1} < x_n$.
Now, $$\frac{1+\dots+x_{n+1}^{n-1}+x_{n+1}^n}{n+1} < \frac{1+\dots+x_{n+1}^{n-1}}{n} < \frac{1+\dots+x_n^{n-1}}{n},$$
where the first inequality comes from the fact that we are dropping the smallest element and therefore the average increases and the second inequality from the fact that we are increasing all elements (except 1).
And as Jose Arnaldo Dris pointed out:
$$
\frac{[x_{n+1}]^{n+1}}{[x_n]^n} = \frac{\frac{1 + x_n + \ldots +  x_n^{n - 1}}{n}}{\frac{1 + x_{n+1} + \ldots + x_{n+1}^n}{n+1}},
$$
which is strictly greater than $1$ as we saw above.
