Suppose $H,C$ are positive semi-definite matrices, and $C$ has unit trace. $H$ is given and fixed while $C$ is allowed to vary. How do I bound the following quantity? $$R=\frac{(\operatorname{Tr} H C)^2}{\operatorname{Tr} H^2 C}$$
For instance suppose
$$H=\left( \begin{array}{cccc} 4 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right)$$
Simulation suggests the following:
$$\frac{16}{25}\le R\le 1$$
Is there a general formula in terms of $H$ eigenvalues?
The ratio $R$ is not always defined as the denominator may be zero. You may modify your question by asking what are the largest possible $k\ge0$ and the least possible $K\ge0$ such that $$ k\operatorname{tr}(H^2C)\le\left(\operatorname{tr}(HC)\right)^2\le K\operatorname{tr}(H^2C) $$ for all $C,H\ge0$ with $\operatorname{tr}(C)=1$.
In this case, the answer is $k=0$ (unless the matrices are $1\times1$, in which case $k=1$) and $K=1$. That $K=1$ is an easy consequence of Cauchy-Schwarz inequality $$ \left(\operatorname{tr}(HC)\right)^2 =\langle H\sqrt{C},\sqrt{C}\rangle_F^2 \le\|H\sqrt{C}\|_F^2\|\sqrt{C}\|_F^2 =\operatorname{tr}(H^2C) $$ and the fact that equality holds when $C=H=\operatorname{diag}(1,0,\ldots,0)$. For the largest possible value of $k$, consider $C=\operatorname{diag}(t,1-t,0,\ldots,0)$ and $H=\operatorname{diag}(1,0,0,\ldots,0)$ with $t>0$. Then $\dfrac{\left(\operatorname{tr}(HC)\right)^2}{\operatorname{tr}(H^2C)}=t$. Since this ratio approaches zero as $t\to0$, we must have $k=0$.
Edit. If you want a lower bound that depends on $H$, an obvious one is $$ \lambda_\min(H)^2\operatorname{tr}(H^2C)\le\lambda_\max(H)^2\left(\operatorname{tr}(HC)\right)^2. $$ To prove it, assume that $H$ is a diagonal matrix. Let $h_\min,h_\max$ and $h_i$ be respectively the minimum, the maximum and the $i$-th diagonal element of $H$. Then $$ h_\min^2\sum_ih_i^2c_{ii} \le h_\min^2h_\max^2\sum_ic_{ii} =h_\max^2\left(\sum_ih_\min c_{ii}\right)^2 \le h_\max^2\left(\sum_ih_ic_{ii}\right)^2. $$
