Which matrix satisfies the following condition? $P$ 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 $P$.
