# Uniform bound on eigenvalues of matrix product

Let $$\Delta:=\lbrace p\in(0,1)^{n}\ \colon\ p_{1}+\dots+p_{n}=1 \rbrace$$ and $$H:=\lbrace h\in\mathbb{R}^{n}\ \colon\ h_{1}+\dots+h_{n}=0 \rbrace$$. Now let $$A\in\mathbb{R}^{n\times n}$$ be a conditionally negative definite matrix, i.e. $$\begin{equation} h\cdot Ah<0,\quad \text{for all }h\in H. \end{equation}$$ It is well-known that for such a matrix there exists $$\lambda>0$$ such that $$\begin{equation} h\cdot Ah\le -\lambda\|h\|^{2},\quad \text{for all }h\in H. \end{equation}$$ Next, consider the matrix field $$D(p)$$, where for $$p\in\Delta$$ $$\begin{equation} D_{ij}(p):=\frac{\delta_{ij}}{p_{i}}-\frac{1}{np_{j}}. \end{equation}$$ Clearly, $$D$$ is conditionally positive definite. In fact it is even 'conditionally elliptic' in that $$\begin{equation} h\cdot D(p)h\ge \|h\|^{2},\quad \text{for all }h\in H \text{ and }p\in\Delta. \end{equation}$$ It it straight-forward to see that for all $$p\in\Delta$$ the product $$AD(p)$$ is again conditionally negative definite. The question which is bothering me is whether we can also get a uniform bound for the quadratic form, i.e. does there exist some $$\rho>0$$ for which $$\begin{equation} h\cdot AD(p)h\le -\rho\|h\|^{2},\quad \text{for all }h\in H\text{ and }p\in\Delta? \end{equation}$$ I appreciate any help.