Let $\mathbb{S}^{dn} \ni X \succeq 0$ with $d,n \in \mathbb{N}$, where $X \succeq 0$ indicates that $X$ is positive semidefinite. Now partition $X$ into the block form \begin{gather} \begin{pmatrix} X_{11} & X_{12} & \cdots & X_{1d} \\ * & X_{22} & \cdots & X_{2d} \\ \vdots & \vdots & \ddots & \vdots \\ * & * & \cdots & X_{dd} \\ \end{pmatrix}, \; \; X_{ii} \in \mathbb{S}^{n} \;\; \text{and} \;\; X_{i,j} \in \mathbb{R}^{n \times n} \;\; \text{with}\;\; i \neq j \end{gather} where the notation $*$ represents the symmetric blocks. Does the matrix inequality \begin{equation} \mathbb{S}^{dn} \ni I_d \otimes \left( \sum_{i=1}^d X_{ii} \right) = \begin{pmatrix} \sum_{i=1}^d X_{ii} & 0_{n\times n} & \cdots & 0_{n\times n} \\ 0_{n\times n} & \sum_{i=1}^d X_{ii} & \cdots & 0_{n\times n} \\ \vdots & \vdots & \ddots & \vdots \\ 0_{n\times n} & 0_{n\times n} & \cdots & \sum_{i=1}^d X_{ii} \\ \end{pmatrix} \succeq X \end{equation} hold for any $\mathbb{S}^{dn} \ni X \succeq 0$?
I believe the above matrix inequality might be considered as an extension of the trace inequality $\mathsf{tr}(X)I_n \succeq X$ with $\mathbb{S}^{n} \ni X \succeq 0$. That is why I believe the inequality is true, however I am not able to give a proof for it.
A followed-up question:
Thank you for the suggestions, indeed, the above inequality is not ture. However, I would like to have a follow-up question which is in fact the original property I intended to prove.
Given matrices $X := \begin{bmatrix} X_1 & X_2 & \cdots & X_d \end{bmatrix} \in \mathbb{R}^{ n \times d \rho n} $ and $\mathbb{S}^{n} \ni U \succ 0$ with $n;d; \rho \in \mathbb{N}$, where the notation $\succ$ standards for the positive definite relation. Let $ \widehat{X} := \begin{bmatrix} X_1 \\ X_2 \\ \vdots \\ X_d \end{bmatrix} \in \mathbb{R}^{dn \times \rho n}$, can we prove \begin{equation} I_d \otimes \left( \widehat{X}^\top \left( I_d \otimes U \right) \widehat{X} \right) \succeq X^\top UX \end{equation} for any given $X \in \mathbb{R}^{ n \times d \rho n}$ and $U \succ 0$?
