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I have found the proof by mathematical induction. Thanks to Sean Shih and Chu-Lan Kao for fruitful discussions.

For $n=1$, the inequality is obvious.

Suppose that for $n=m\in\mathcal{N}$, $\det(\mathbf{D}\mathbf{D}^H+\mathbf{R}\mathbf{R}^H)\geq \prod_{i=1}^m(|\mathbf{D}_{ii}|^2+|\mathbf{R}_{ii}|^2)$ holds.

For $n = m+1$, let $\mathbf{D}'=\left[\begin{matrix}\mathbf{D} & 0\\ 0 & d\end{matrix}\right]$ and $\mathbf{R}'=\left[\begin{matrix}\mathbf{R} & \mathbf{r}\\ \mathbf{0} & r\end{matrix}\right]$ where $\mathbf{D}$ is an $m\times m$ diagonal matrix, $\mathbf{R}$ is an $m\times m$ upper-triangular matrix, $\mathbf{r}$ is an $m \times 1$ vector, and $d$ and $r$ are scalars. Let $\mathbf{A} = \mathbf{D}\mathbf{D}^H+\mathbf{R}\mathbf{R}^H$ be non-singular, or the inequality is obvious. Then \begin{align} &\det(\mathbf{D}'\mathbf{D}'^H+\mathbf{R}'\mathbf{R}'^H) \\ = &\det(\mathbf{A}+\mathbf{rr}^H)(|d|^2+|r|^2-|r|^2\mathbf{r}^2(\mathbf{A}+\mathbf{rr}^H)^{-1}\mathbf{r}^H)\\ =&\det(\mathbf{A})(1+\mathbf{r}^H\mathbf{A}^{-1}\mathbf{r})(|d|^2+|r|^2-|r|^2(1-(1+\mathbf{r}^H\mathbf{A}^{-1}\mathbf{r})^{-1}))\\ =&\det(\mathbf{A})(|d|^2+|r|^2)+\det(\mathbf{A})|d|^2\mathbf{r}^H\mathbf{A}^{-1}\mathbf{r}\\ \geq &\det(\mathbf{A})(|d|^2+|r|^2) = \prod_{i=1}^{m+1}(|\mathbf{D}'_{ii}|^2+|\mathbf{R}'_{ii}|^2). \end{align} By mathematical induction, the proof is completed.

I have found the proof by mathematical induction. Thanks to Sean Shih and Chu-Lan Kao for fruitful discussions.

For $n=1$, the inequality is obvious.

Suppose that for $n=m\in\mathcal{N}$, $\det(\mathbf{D}\mathbf{D}^H+\mathbf{R}\mathbf{R}^H)\geq \prod_{i=1}^m(|\mathbf{D}_{ii}|^2+|\mathbf{R}_{ii}|^2)$ holds.

For $n = m+1$, let $\mathbf{D}'=\left[\begin{matrix}\mathbf{D} & 0\\ 0 & d\end{matrix}\right]$ and $\mathbf{R}'=\left[\begin{matrix}\mathbf{R} & \mathbf{r}\\ \mathbf{0} & r\end{matrix}\right]$ where $\mathbf{D}$ is an $m\times m$ diagonal matrix, $\mathbf{R}$ is an $m\times m$ upper-triangular matrix, $\mathbf{r}$ is an $m \times 1$ vector, and $d$ and $r$ are scalars. Let $\mathbf{A} = \mathbf{D}\mathbf{D}^H+\mathbf{R}\mathbf{R}^H$ be non-singular, or the inequality is obvious. Then \begin{align} &\det(\mathbf{D}'\mathbf{D}'^H+\mathbf{R}'\mathbf{R}'^H) \\ = &\det(\mathbf{A}+\mathbf{rr}^H)(|d|^2+|r|^2-|r|^2\mathbf{r}^2(\mathbf{A}+\mathbf{rr}^H)^{-1}\mathbf{r}^H)\\ =&\det(\mathbf{A})(1+\mathbf{r}^H\mathbf{A}^{-1}\mathbf{r})(|d|^2+|r|^2-|r|^2(1-(1+\mathbf{r}^H\mathbf{A}^{-1}\mathbf{r})^{-1}))\\ =&\det(\mathbf{A})(|d|^2+|r|^2)+\det(\mathbf{A})|d|^2\mathbf{r}^H\mathbf{A}^{-1}\mathbf{r}\\ \geq &\det(\mathbf{A})(|d|^2+|r|^2) = \prod_{i=1}^{m+1}(|\mathbf{D}'_{ii}|^2+|\mathbf{R}'_{ii}|^2) \end{align} where we use matrix inversion lemma and matrix determinant lemma. By mathematical induction, the proof is completed.

I have found the proof by mathematical induction. Thanks to Sean Shih and Chu-Lan Kao for fruitful discussions.

For $n=1$, the inequality is obvious.

For $n=2$, let $\mathbf{D} = \left[\begin{matrix}d_{1} & 0\\ 0 & d_2\end{matrix}\right]$ and $\mathbf{R} = \left[\begin{matrix}r_{11} & r_{12}\\ 0 & r_{22}\end{matrix}\right]$. Then \begin{align} \det(\mathbf{D}\mathbf{D}^H+\mathbf{R}\mathbf{R}^H) &= \det\left(\left[\begin{matrix}|d_1|^2+|r_{11}|^2+|r_{12}|^2 & r_{12}r_{21}^*\\ r_{22}r_{12}^* & |d_2|^2+|r_{22}|^2\end{matrix}\right]\right)\\ &\geq (|d_1|^2+|r_{11}|^2)(|d_2|^2+|r_{22}|^2). \end{align}

Suppose that for $n=m\in\mathcal{N}$, $\det(\mathbf{D}\mathbf{D}^H+\mathbf{R}\mathbf{R}^H)\geq \prod_{i=1}^m(|\mathbf{D}_{ii}|^2+|\mathbf{R}_{ii}|^2)$ holds.

For $n = m+1$, let $\mathbf{D}'=\left[\begin{matrix}\mathbf{D} & 0\\ 0 & d\end{matrix}\right]$ and $\mathbf{R}'=\left[\begin{matrix}\mathbf{R} & \mathbf{r}\\ \mathbf{0} & r\end{matrix}\right]$ where $\mathbf{D}$ is an $m\times m$ diagonal matrix, $\mathbf{R}$ is an $m\times m$ upper-triangular matrix, $\mathbf{r}$ is an $m \times 1$ vector, and $d$ and $r$ are scalars. Let $\mathbf{A} = \mathbf{D}\mathbf{D}^H+\mathbf{R}\mathbf{R}^H$ be non-singular, or the inequality is obvious. Then \begin{align} &\det(\mathbf{D}'\mathbf{D}'^H+\mathbf{R}'\mathbf{R}'^H) \\ = &\det(\mathbf{A}+\mathbf{rr}^H)(|d|^2+|r|^2-|r|^2\mathbf{r}^2(\mathbf{A}+\mathbf{rr}^H)^{-1}\mathbf{r}^H)\\ =&\det(\mathbf{A})(1+\mathbf{r}^H\mathbf{A}^{-1}\mathbf{r})(|d|^2+|r|^2-|r|^2(1-(1+\mathbf{r}^H\mathbf{A}^{-1}\mathbf{r})^{-1}))\\ =&\det(\mathbf{A})(|d|^2+|r|^2)+\det(\mathbf{A})|d|^2\mathbf{r}^H\mathbf{A}^{-1}\mathbf{r}\\ \geq &\det(\mathbf{A})(|d|^2+|r|^2) = \prod_{i=1}^{m+1}(|\mathbf{D}'_{ii}|^2+|\mathbf{R}'_{ii}|^2). \end{align} By mathematical induction, the proof is completed.

I have found the proof by mathematical induction. Thanks to Sean Shih and Chu-Lan Kao for fruitful discussions.

For $n=1$, the inequality is obvious.

Suppose that for $n=m\in\mathcal{N}$, $\det(\mathbf{D}\mathbf{D}^H+\mathbf{R}\mathbf{R}^H)\geq \prod_{i=1}^m(|\mathbf{D}_{ii}|^2+|\mathbf{R}_{ii}|^2)$ holds.

For $n = m+1$, let $\mathbf{D}'=\left[\begin{matrix}\mathbf{D} & 0\\ 0 & d\end{matrix}\right]$ and $\mathbf{R}'=\left[\begin{matrix}\mathbf{R} & \mathbf{r}\\ \mathbf{0} & r\end{matrix}\right]$ where $\mathbf{D}$ is an $m\times m$ diagonal matrix, $\mathbf{R}$ is an $m\times m$ upper-triangular matrix, $\mathbf{r}$ is an $m \times 1$ vector, and $d$ and $r$ are scalars. Let $\mathbf{A} = \mathbf{D}\mathbf{D}^H+\mathbf{R}\mathbf{R}^H$ be non-singular, or the inequality is obvious. Then \begin{align} &\det(\mathbf{D}'\mathbf{D}'^H+\mathbf{R}'\mathbf{R}'^H) \\ = &\det(\mathbf{A}+\mathbf{rr}^H)(|d|^2+|r|^2-|r|^2\mathbf{r}^2(\mathbf{A}+\mathbf{rr}^H)^{-1}\mathbf{r}^H)\\ =&\det(\mathbf{A})(1+\mathbf{r}^H\mathbf{A}^{-1}\mathbf{r})(|d|^2+|r|^2-|r|^2(1-(1+\mathbf{r}^H\mathbf{A}^{-1}\mathbf{r})^{-1}))\\ =&\det(\mathbf{A})(|d|^2+|r|^2)+\det(\mathbf{A})|d|^2\mathbf{r}^H\mathbf{A}^{-1}\mathbf{r}\\ \geq &\det(\mathbf{A})(|d|^2+|r|^2) = \prod_{i=1}^{m+1}(|\mathbf{D}'_{ii}|^2+|\mathbf{R}'_{ii}|^2). \end{align} By mathematical induction, the proof is completed.

I have found the proof by mathematical induction. Thanks to Sean Shih and Chu-Lan Kao for fruitful discussions.

For $n=1$, the inequality is obvious.

For $n=2$, let $\mathbf{D} = \left[\begin{matrix}d_{1} & 0\\ 0 & d_2\end{matrix}\right]$ and $\mathbf{R} = \left[\begin{matrix}r_{11} & r_{12}\\ 0 & r_{22}\end{matrix}\right]$. Then $\det(\mathbf{D}\mathbf{D}^H+\mathbf{R}\mathbf{R}^H) = \det\left(\left[\begin{matrix}|d_1|^2+|r_{11}|^2+|r_{12}|^2 & r_{12}r_{21}^*\\ r_{22}r_{12}^* & |d_2|^2+|r_{22}|^2\end{matrix}\right]\right)\geq (|d_1|^2+|r_{11}|^2)(|d_2|^2+|r_{22}|^2)$.

Suppose that for $n=m\in\mathcal{N}$, $\det(\mathbf{D}\mathbf{D}^H+\mathbf{R}\mathbf{R}^H)\geq \prod_{i=1}^m(|\mathbf{D}_{ii}|^2+|\mathbf{R}_{ii}|^2)$ holds.

For $n = m+1$, let $\mathbf{D}'=\left[\begin{matrix}\mathbf{D} & 0\\ 0 & d\end{matrix}\right]$ and $\mathbf{R}'=\left[\begin{matrix}\mathbf{R} & \mathbf{r}\\ \mathbf{0} & r\end{matrix}\right]$ where $\mathbf{D}$ is an $m\times m$ diagonal matrix, $\mathbf{R}$ is an $m\times m$ upper-triangular matrix, $\mathbf{r}$ is an $m \times 1$ vector, and $d$ and $r$ are scalars. Let $\mathbf{A} = \mathbf{D}\mathbf{D}^H+\mathbf{R}\mathbf{R}^H$ be non-singular, or the inequality is obvious. Then \begin{align} &\det(\mathbf{D}'\mathbf{D}'^H+\mathbf{R}'\mathbf{R}'^H) \\ = &\det(\mathbf{A}+\mathbf{rr}^H)(|d|^2+|r|^2-|r|^2\mathbf{r}^2(\mathbf{A}+\mathbf{rr}^H)^{-1}\mathbf{r}^H)\\ =&\det(\mathbf{A})(1+\mathbf{r}^H\mathbf{A}^{-1}\mathbf{r})(|d|^2+|r|^2-|r|^2(1-(1+\mathbf{r}^H\mathbf{A}^{-1}\mathbf{r})^{-1}))\\ =&\det(\mathbf{A})(|d|^2+|r|^2)+\det(\mathbf{A})|d|^2\mathbf{r}^H\mathbf{A}^{-1}\mathbf{r}\\ \geq &\det(\mathbf{A})(|d|^2+|r|^2) = \prod_{i=1}^{m+1}(|\mathbf{D}'_{ii}|^2+|\mathbf{R}'_{ii}|^2). \end{align} By mathematical induction, the proof is completed.

I have found the proof by mathematical induction. Thanks to Sean Shih and Chu-Lan Kao for fruitful discussions.

For $n=1$, the inequality is obvious.

For $n=2$, let $\mathbf{D} = \left[\begin{matrix}d_{1} & 0\\ 0 & d_2\end{matrix}\right]$ and $\mathbf{R} = \left[\begin{matrix}r_{11} & r_{12}\\ 0 & r_{22}\end{matrix}\right]$. Then \begin{align} \det(\mathbf{D}\mathbf{D}^H+\mathbf{R}\mathbf{R}^H) &= \det\left(\left[\begin{matrix}|d_1|^2+|r_{11}|^2+|r_{12}|^2 & r_{12}r_{21}^*\\ r_{22}r_{12}^* & |d_2|^2+|r_{22}|^2\end{matrix}\right]\right)\\ &\geq (|d_1|^2+|r_{11}|^2)(|d_2|^2+|r_{22}|^2). \end{align}

Suppose that for $n=m\in\mathcal{N}$, $\det(\mathbf{D}\mathbf{D}^H+\mathbf{R}\mathbf{R}^H)\geq \prod_{i=1}^m(|\mathbf{D}_{ii}|^2+|\mathbf{R}_{ii}|^2)$ holds.

For $n = m+1$, let $\mathbf{D}'=\left[\begin{matrix}\mathbf{D} & 0\\ 0 & d\end{matrix}\right]$ and $\mathbf{R}'=\left[\begin{matrix}\mathbf{R} & \mathbf{r}\\ \mathbf{0} & r\end{matrix}\right]$ where $\mathbf{D}$ is an $m\times m$ diagonal matrix, $\mathbf{R}$ is an $m\times m$ upper-triangular matrix, $\mathbf{r}$ is an $m \times 1$ vector, and $d$ and $r$ are scalars. Let $\mathbf{A} = \mathbf{D}\mathbf{D}^H+\mathbf{R}\mathbf{R}^H$ be non-singular, or the inequality is obvious. Then \begin{align} &\det(\mathbf{D}'\mathbf{D}'^H+\mathbf{R}'\mathbf{R}'^H) \\ = &\det(\mathbf{A}+\mathbf{rr}^H)(|d|^2+|r|^2-|r|^2\mathbf{r}^2(\mathbf{A}+\mathbf{rr}^H)^{-1}\mathbf{r}^H)\\ =&\det(\mathbf{A})(1+\mathbf{r}^H\mathbf{A}^{-1}\mathbf{r})(|d|^2+|r|^2-|r|^2(1-(1+\mathbf{r}^H\mathbf{A}^{-1}\mathbf{r})^{-1}))\\ =&\det(\mathbf{A})(|d|^2+|r|^2)+\det(\mathbf{A})|d|^2\mathbf{r}^H\mathbf{A}^{-1}\mathbf{r}\\ \geq &\det(\mathbf{A})(|d|^2+|r|^2) = \prod_{i=1}^{m+1}(|\mathbf{D}'_{ii}|^2+|\mathbf{R}'_{ii}|^2). \end{align} By mathematical induction, the proof is completed.