If we consider the problem in a spherically symmetric potential, we can see that with an increase in the orbital quantum number, the energy state in the spectrum increases. This is observed both in the hydrogen atom and in the spherical well. Can we somehow prove that this is so in any $\rm SO(3)$ symmetric problem?
Equation for this:
$$ \left( -\dfrac{1}{2 m}\dfrac{\partial^2}{\partial r^2} + \dfrac{\ell(\ell+1)}{2 m r^2} + V(r) \right) R(r)= E_{n,\ell} R(r) $$
Your assumptions are, unfortunately, about as wrong as they can be.
we can see that with an increase in the orbital quantum number, the energy state in the spectrum increases
This is false in general.
This is observed both in the hydrogen atom and [the isotropic harmonic oscillator]
This is doubly false: you've managed to get precisely the two examples where the behaviour you want doesn't happen.
To be more specific, if you consider hamiltonians of the form $H = \frac12 \mathbf p^2 + C r^n$ in three dimensions, where $n$ is some power and $C$ is a constant (with $\mathrm{sgn}(C)\cdot\mathrm{sgn}(n)>0$ to ensure the existence of bound states), then the spectrum is degenerate in the magnetic quantum number $m$ (because of the rotational invariance), but the energies $E_{n\ell}$ will generally depend on the eigenvalue $\ell$ of the angular momentum. This has precisely two exceptions, namely the hydrogen atom and the isotropic harmonic oscillator, at $n=-1$ and $n=2$ respectively.
In the hydrogen atom, this is reflected in the fact that the different subshells within a given-$n$ shell (so e.g. the $3s$, $3p$ and $3d$ subshells) have the same energy (ignoring fine-structure contributions). This is essentially because the system is more symmetric than it appears: it is invariant under a copy of $\rm SO(4)$, not just $\rm SO(3)$, which leads to the conservation of the Runge-Lenz vector and to the degeneracy in the $\ell$ shells.
In the isotropic harmonic oscillator, there is a similar additional symmetry and a similar additional conserved quantity (the so-called Fradkin tensor) which ensure degeneracy between different $\ell$s. As a simple example, consider the states $(x^2+y^2-1) e^{-(x^2+y^2+z^2)/\sigma^2}$ and $(x+iy)^2 e^{-(x^2+y^2+z^2)/\sigma^2}$, with $\ell=0$ and $\ell=2$, and which can be shown to be superpositions of the tensor-product wavefunctions $\psi_2(x)\psi_0(y)\psi_0(z)$, $\psi_0(x)\psi_2(y)\psi_0(z)$ and $\psi_1(x)\psi_1(y)\psi_0(z)$ when working on the separable 3×1D basis.
In both cases, the additional symmetry is related to the fact that the $n=-1$ and $n=2$ cases are the only ones to support closed orbits.
Also, in both cases, you can go out of your way to define "radial" quantum numbers (basically $n_r=n-\ell$) such that if you keep $n_r$ constant and increase $\ell$ then $E_{n\ell}$ increases. This is entirely a red herring: if you increase $n$ the energy obviously increases, but you're doing completely the wrong comparison.
More generally, if you want to
somehow prove that this is so in any $\rm SO(3)$ symmetric problem?
you need only consider a hamiltonian of the form $$ H = H_0 - \alpha L^2, $$ where $H_0$ is any spherically-symmetric hamiltonian (say, one of the two degenerate examples from above) and $\alpha$ is large enough, to get a system for which increasing $\ell$ gives you an identical eigenbasis but with reduced energy eigenvalues on all or a substantial subset of that eigenbasis. This then acts as a clear counterexample to your claim.
Finally, at this point, the last refuge the claim has is to go into something like
OK, suppose we just consider hamiltonians of the form $H = \frac12 \mathbf p^2 + V(r)$, where $V(r)$ is real-valued (and we ignore a bunch of other possible $\rm SO(3)$-invariant terms like $f(r)L^2$ or $g(r)L^4$ or the like), and we enforce ordering on the radial quantum number, i.e. insist on just seeing the radial problems $$ \hat H_\ell = \frac12 \hat p_r^2 + \frac{\hbar^2\ell(\ell+1)}{2m \hat r^2} + V(\hat r) $$ (where $\langle r|\hat p_r = \frac1r \frac{\partial}{\partial r}r\langle r|$ for the radial momentum operator) as separate, independent problems. Is the $n_r$th eigenvalue $E_{n_r; \ell}$ of $\hat H_\ell$ an increasing function of $\ell$?
Here the answer is yes for the ground state, which can be shown using the fact that $\hat r^{2}$ (together with its inverse) is a positive-definite operator (so if $\ell_1<\ell_2$ you can provide a bound for the expectation value of $\hat H_{\ell_1}$ at the ground state of $\hat H_{\ell_2}$ which then bounds the ground-state energy of $\hat H_{\ell_1}$).
For the excited states - to be honest, I'm not sure. That narrower question, suitably narrowed down to the simpler version
Given a self-adjoint operator $H$ on $L_2(\mathbb R)$ and a positive-definite change $\Delta H$, is the $k$th eigenvalue of $H+\Delta H$ greater than the $k$th eigenvalue of $H$?
would make a good question for the maths site.
