Why covariance constraint subsumes the average power constraint?

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I am studying an optimization problem in the form of \begin{equation} \begin{aligned} &\underset{p(x)}{\text{maximize}} & & W\\ & \text{subject to} & & 0 \preceq K_{X} \preceq S, \end{aligned} \end{equation} in which $K_{X}$ is the covariance matrix of $X$, and $S$ is a positive semidefinite matrix. The goal is to find optimal distribution (optimal $p(x)$) for that. The paper says that since the covariance constraint subsumes the average power constraint we conclude that the same $p(x)$ is optimal for the following problem too: \begin{equation} \begin{aligned} &\underset{p(x)}{\text{maximize}} & & W\\ & \text{subject to} & & 0 \le \operatorname{tr}(K_{X}) \le P. \end{aligned} \end{equation} Can anyone explain this?

My question is simply what asked in the title, but the following details might help. In that specific problem, $W \triangleq h(X + Z_1) - \mu h(X + Z_2)$ where $Z_1$ and $Z_2$ are independent Gaussian vectors and $\mu\ge 1$ is is an scalar, and the vector $X$ is independent of $Z_1$ and $Z_2$, $h(\cdot)$ is the differential entropy function, and $P$ is a (scalar) constant.