Matrix eigenvalues inequality (2)

Suppose 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 $$m \times m$$ positive matrix, whose eigenvalues are $$b_1\ge b_2\ldots \ge b_m>0;$$ $$C$$ is a $$p \times p$$ positive matrix, whose eigenvalues are $$c_1\ge c_2\ldots \ge c_p>0.$$ For $$n\times p$$ full rank matrix $$X$$ with $$n\ge p,$$ a $$n\times m$$ full rank matrix $$Y$$ with $$n\ge m,$$, and $$m\ge p,$$

Question: How to prove it? $$det\Big(X'(A+YBY')^{-1}X+C\Big)\ge l(X,Y)\prod_{i=1}^p\Big(\frac{1}{a_i+b_{i}}+c_{p-i+1}\Big),$$ where $$l(X,Y)$$ is a positive constant that only depends on $$X,Y.$$

• Let $\lambda_1\geqslant \lambda_{2}\geqslant \ldots \geqslant \lambda_p$ be eigenvalues of $X('A+YBY')X+C$. It suffices to prove that $$\lambda_{p-i+1}\geqslant l(X,Y)(\frac{1}{a_i+b_i}+c_{p-i+1})\quad (*)$$ for all $i=1,\dots,p.$ Consider two cases. 1) $c_{p-i+1}\geqslant \frac{1}{a_i+b_i}$. It is easy to handle it. 2) $\frac{1}{a_i+b_i}> c_{p-i+1}.$ It suffices to prove $\lambda_{p-i+1}\ge l(X,Y) \frac{1}{a_i+b_i}.$ Let $\lambda_1\geqslant v_{2}\geqslant \ldots \geqslant v_p$ be eigenvalues of $X'(A+YBY')^{-1}X$. Feb 2 '19 at 14:57
• Then $\lambda_{p-i+1}\ge v_{p-i+1}.$ By the variational principle, I can prove $v_{p-i+1}\ge l(X,Y) t_i^{-1},$ where $t_1\geqslant t_{2}\geqslant \ldots \geqslant t_n$ be eigenvalues of $A+YBY'$. I don't know how to do next. Because for $i=1,\ldots,p$ $t_i\le l(X,Y) (a_{i}+b_i)$ does not always hold. Feb 2 '19 at 15:08