Assume that $A$ is a $n\times n$ positive matrix, whose eigenvalues are $a_1\ge a_2\ldots \ge a_n>0;$ $B$ is a $p \times p$ positive matrix, whose eigenvalues are $b_1\ge b_2\ldots \ge b_p>0.$ For $n\times p$ full rank matrix $X$ with $n\ge p,$

Question: How to prove $$|X'AX+B|\ge c(X)\prod_{i=1}^p(a_{n-p+i}+b_{i})，$$

where $|\cdot|$ stands for determinant, $c(X)$ is a postive constant, only denpends on $X.$

Assume that $A$ is a $n\times n$ positive matrix, whose eigenvalues are $a_1\ge a_2\ldots \ge a_n>0;$ $B$ is a $p \times p$ positive matrix, whose eigenvalues are $b_1\ge b_2\ldots \ge b_p>0.$ For $n\times p$ full rank matrix $X$ with $n\ge p,$

How to prove $$\det(X'AX+B)\ge c(X)\prod_{i=1}^p(a_{n-p+i}+b_{i})，$$

where $c(X)$ is a postive constant that depends only on $X$.

Assume that $A$ is a $n\times n$ positive matrix, whose eigenvalues are $a_1\ge a_2\ldots \ge a_n>0;$ $B$ is a $p \times p$ positive matrix, whose eigenvalues are $b_1\ge b_2\ldots \ge b_p>0.$ For $n\times p$ full rank matrix $X$ with $n\ge p,$

Question: Is there a similar inequality, such that $$|X'AX+B|\ge c(X)\prod_{i=1}^p(a_i+b_{p-i+1}),$$ or $$|X'AX+B|\ge c(X)\prod_{i=1}^p(a_{p-i+1}+b_i)$$ or something else.

where $|\cdot|$ stands for determinant, $c(X)$ is a postive constant, only denpends on $X.$

Assume that $A$ is a $n\times n$ positive matrix, whose eigenvalues are $a_1\ge a_2\ldots \ge a_n>0;$ $B$ is a $p \times p$ positive matrix, whose eigenvalues are $b_1\ge b_2\ldots \ge b_p>0.$ For $n\times p$ full rank matrix $X$ with $n\ge p,$

Question: How to prove $$|X'AX+B|\ge c(X)\prod_{i=1}^p(a_{n-p+i}+b_{i})，$$

where $|\cdot|$ stands for determinant, $c(X)$ is a postive constant, only denpends on $X.$