$P$ is a real (symmetric) positive definite matrix. Let $P_i$ and $P_j$ represent the $i$'th and $j$'th columns of $P$, respectively. Further, let $P_{ki}$ represent the element situated at the $k$'th row of the column vector $P_i$.
I want to find additional conditions on the matrix $P$ such that the following inequality holds for all $i,j,k$:
$$ 0 \leq \frac{P_{ij} P_{kj}}{(c + P_{jj})} \leq 2 P_{ki}, $$
with $c > 0$ and $P_{ij}$ represents the $(i,j)$'th element of $P$.
If $P$ is strictly ultrametric, then, would the above inequality be satisfied?
$P$ is a strictly ultrametric matrix of size $n \times n$ if:
- All the elements of $P$ are non-negative
- $\forall (i,j, i \neq j) \in (1,\dots,n): P(i,i) > P(i,j)$
- $\forall (i,j,k) \in (1,\dots,n): P(i,j) \geq {\rm{min}}(P(i,k),P(k,j))$
Any logical conjectures would be appreciated if a direct answer is hard to come by.
Let $A$ be the following matrix $a_{ik} = \log_2 P_{ik}$. Then $$ a_{ij} + a_{kj} \leq 1 + a_{ik} + a_{jj} \tag{*} $$ is sufficient for $$ \frac{P_{ij} P_{kj}}{c + P_{jj}} < \frac{P_{ij} P_{kj}}{P_{jj}} \leq 2P_{ik}. $$
If $a_{ik} > -1, a_{jj} > 0$ and the matrix $A$ is diagonally dominant $|a_{jj}| \geq \sum_{k=1}^n |a_{kj}|$ then $$ a_{jj} = |a_{jj}| \geq \sum_{k=1}^n |a_{kj}| \geq |a_{ij}| + |a_{kj}| \geq a_{ij} + a_{kj}. $$
There are also weaker conditions that guarantee $(*)$. If the matrix $A$ is pseudo-ultrametric in the following sense $$ \begin{gather} a_{ik} \geq \min(a_{ij}, a_{kj}) \tag{1}\\ a_{jj} + 1 \geq \max_s a_{sj} \geq \max(a_{ij}, a_{kj}).\tag{2} \end{gather} $$ Summing (1) and (2) yields $$ 1 + a_{ik} + a_{jj} \geq \min(a_{ij}, a_{kj}) + \max(a_{ij}, a_{kj}) = a_{ij} + a_{kj}. $$
You can easily reformulate the conditions in terms of the original matrix $P$.