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 $X$.