Is the trace of a product of positive definite matrices bigger than the sum of the product of their corresponding diagonal elements?

Question

Let $\mathbf{S},\mathbf{\Theta}\in\mathbb{R}^{p\times p}$ by symmetric, positive definite matrices. Denote their corresponding $(i,j)$-th component by $S_{ij},\Theta_{ij}$. Is it true that \begin{align*} \text{Trace}(\mathbf{S\Theta})\geq\sum_{i=1}^{p}S_{ii}\Theta_{ii}. \end{align*}

Answer 1

This is not true. Take $$A= \begin{bmatrix} 2& 1 \\ 1 & 2 \end{bmatrix}$$ and $$B= \begin{bmatrix} 2& -1 \\ -1 & 2 \end{bmatrix}$$

Answer 2

No. If it is true, by a continuity argument, the inequality should still be true when $S$ and $\Theta$ are positive semidefinite, but it is not: $$ S=\pmatrix{1&-1\\ -1&1}, \ \Theta=\pmatrix{1&1\\ 1&1} \ \Rightarrow\ \operatorname{tr}(S\Theta)=0<2=\sum_iS_{ii}\Theta_{ii}. $$