Can we get closed-form solutions for this optimization problem? The optimization problem is as follows:
\begin{equation}
\begin{aligned}
&\max\limits_{\boldsymbol{\omega}} \frac{\sum\limits_{n=1}^N \frac{\omega_n k_n}{k_n-1}}{\sqrt{\sum\limits_{n=1}^N \omega_n^2 \frac{k_n^3}{(k_n-1)^2(k_n-2)}}}\\
&\text{s.t.}\sum\limits_{n=1}^N\omega_n=1
\end{aligned}
\end{equation}
where $\boldsymbol{\omega}=\left[\omega_1,\omega_2,\cdots,\omega_N \right], 0\leq\omega_n\leq1$, and $k_n$ are integers such that $k_1 \neq k_2\cdots \neq k_N$.
Problem:
Can we get closed-from solutions for this optimization problem?
 A: Too long for a comment.
Calling $a_i = \frac{k_i}{k_i-1}$ and $b_i = \frac{k_i^3}{(k_i-1)^2(k_i-2)}$ the problem is equivalent to
$$
\max_{\omega}\frac{(a\cdot\omega)^2}{\sum b_i\omega_i^2}\ \ \text{s. t.}\ \ \cases{\sum \omega_i = 1\\ \ 0\le\omega_i\le 1}
$$
The lagrangian
$$
L(\omega,\lambda) = \frac{(a\cdot\omega)^2}{\sum b_i\omega_i^2}+\lambda\left(\sum \omega_i - 1\right)
$$
has stationary points at
$$
\cases{
2\frac{(a\cdot\omega)a_i}{\sum b_i\omega_i^2}-2\frac{(a\cdot\omega)^2\omega_i b_i}{\left(\sum b_i\omega_i^2\right)^2}+\lambda=0\\
\sum \omega_i = 1}
$$
or
$$
\cases{
(\sum b_i\omega_i^2)a_i-(a\cdot\omega)\omega_i b_i+\frac 12\frac{(\sum b_i\omega_i^2)^2}{a\cdot\omega}\lambda=0\\
\sum \omega_i = 1}
$$
or as $\lambda$ is generic, we follow with $\mu = \frac 12\frac{(\sum b_i\omega_i^2)^2}{a\cdot\omega}\lambda$
$$
\cases{
(\sum b_i\omega_i^2)a_i-(a\cdot\omega)\omega_i b_i+\mu=0\\
\sum \omega_i = 1}
$$
now multiplying the line $i$ by $\omega_i$ and adding all lines we can conclude that $\mu = 0$
NOTE
This implies on
$$
\frac{\omega_i b_i}{a_i}=\kappa,\ \forall i
$$
and finally
$$
\omega_i = \frac{\frac{a_i}{b_i}}{\sum\frac{a_i}{b_i}}
$$
so we have also $0\le \omega_i\le 1,\ \ \forall i$
