Matrix eigenvalues inequality (1) 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 $X$. 
 A: 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 $\lambda_i\geqslant 2\rho(X)a_{n-p+i}\geqslant \rho(a_{n-p+i}+b_i)$.
