Given a positive definite matrix $A$ there have to be some restrictions on the off-diagonal elements $A_{ij}$ (i.e. where $i \ne j$). Since there exists a Cholesky decomposition (i.e. $A=U^TU$) for a positive definite matrix I know that in a 2x2 matrix $U_{22} = \sqrt{A_{22} - \frac{A_{12}^{2}}{A_{11}}}$ from which I can derive that $A_{12} \lt \sqrt{A_{11}A_{22}}$ as otherwise the value for $U_{22}$ would be complex and $A=U^TU$ holds no longer true.
When I increase the dimensions of the my matrix $A$ to 3x3 or 4x4, the cholesky decomposition gets longer but is still manageable but using something larger is something I can no longer handle.
My question is if there are also other methods to impose restrictions on the off-diagonal elements of a positive definite matrix other that are easier to handle than the cholesky decomposition?
The equivalency $xAx = \sum_{ij} A_{ij} x_i x_j$ gives a direct way to bound the off-diagonal elements relative to the diagonal for a real symmetric positive definite matrix.
In general, $|A_{ij}| < (A_{ii} + A_{jj}) / 2$.
To see this, set $x$ equal to a vector with a value 1 at positions $i$ and $j$, and zero elsewhere.
We then have $A_{ii} + 2A_{ij} + A_{jj} > 0$. This implies the negative side of the bound $-A_{ij} < (A_{ii} + A_{jj}) / 2$.
Using a vector $x$ with 1 at position $i$ and $-1$ at position $j$, we get the positive side of the bound.
This gives us the bound on the absolute value of any off-diagonal element.
You can also obtain a bound based on the geometric mean of $A_{ii}$ and $A_{jj}$ (and an extension of the bound you have above) using 1 at position $i$ and $\lambda$ at position $j$ using the method at https://math.stackexchange.com/a/3018652/1175270.
This gives $|A_{ij}| < \sqrt{A_{ii} A_{jj}}$.