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 $n\ge p,$



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> How to prove
>  $$\det(X'AX+B)\ge c(X)\prod\_{i=1}^p(a\_{n-p+i}+b\_{i})，$$
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where $c(X)$ is a postive constant that depends only on $X$. 



Matrix eigenvalues inequality (1)
Your both conjectural inequalities are equivalent to each other and false even for $n=p$, $A=B$, $X=I$ (and $a\_i=\lambda^i$ for large $\lambda$, for example).


What is true that $$|X'AX+B|\geqslant c(X)\prod\_{i=1}^p (a\_{n-p+i}+b\_i).$$


**Proof.** Let $\lambda\_1\geqslant \lambda\_{2}\geqslant \ldots \geqslant \lambda\_p$ be eigenvalues of $X'AX+B$. It suffices to prove that $$\lambda\_{i}\geqslant \rho(b\_i+a\_{n-p+i})\quad (\*)$$ for all $i=1,\dots,p$, where $\rho=\rho(X)$ depends only on $X$. 
Consider two cases.


1) $b\_i\geqslant a\_{n-p+i}$. Then $(\*)$ holds with $\rho=1/2$, since $X'AX+B\geqslant B$, thus $\lambda\_i\geqslant b\_i$.


2) $a\_{n-p+i}\geqslant b\_i$. Let $L$ be the image of the operator (identified with the matrix) $X$, $\dim L=p$ and $X$ is a linear isomorphism onto $L$. Denote by $4\rho^2$, $\rho>0$, the norm of the inverse map $X^{-1}:L\mapsto X$. The intersection $L\_i$ of $L$ and the space generated by $u\_1,\dots,u\_{n-p+i}$, where $Au\_i=a\_i u\_i$ and $u\_i$ are orthogonal eigenvectors of $A$, has dimension at least $i$. We have $$((X'AX+B) x,x)\geqslant (X'AX x,x)=(AXx,Xx)\geqslant a\_{n-p+i}(Xx,Xx)\geqslant 2\rho a\_{n-p+i}(x,x)$$ whenever $x\in X^{-1}L\_i.$ Therefore by the variational (Courant) principle $\lambda\_i\geqslant 2\rho(X)a\_{n-p+i}\geqslant \rho(a\_{n-p+i}+b\_i)$.


