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Singular value or eigenvalue problems lie at the center of matrix analysis. One classical result is
$$\lambda_{j}(X^{*}X+Y^{*}Y)\geq 2\sigma_j(XY^*)$$
for $j \in \{1, \ldots, n\}$, where $\lambda_j(\cdot)$ and $\sigma_j(\cdot)$ denote the $j$th largest eigenvalue and singular value, respectively, and $X$ and $Y$ are $n \times n$ complex matrices. I conjecture that the following does hold.
$$\lambda_{j}((I+X^{*}X)(I+Y^{*}Y))\geq\sigma_j^2(I+XY^*)$$
The conjecture is true.
Lemma 1 : For every matrix $Z$ holds $\lambda_j(Z^* Z + Z^* + Z ) \leq \lambda_j(Z^* Z + 2 (Z^* Z)^{1/2})$ .
For a proof see the proof of $\lambda_j(Z^* + Z ) \leq \lambda_j(2 (Z^* Z)^{1/2})$ in Bhatia, Matrix Analysis, Proposition III.5.1 (Fan-Hoffman).
Lemma 2 : For $a \geq 0 , b > 0$ holds $(a+b)(1+b^{-1}) \geq a + 2 a^{1/2} + 1$ .
Proof left to the reader.
Proof of the conjecture :
Choose $Z = XY^*$, $A=Z^* Z$, $B=YY^*$ .
We may assume that $B$ is invertable.
Then we have to show :
$\lambda_j((I+B^{-1})^{1/2}(A+B)(I+B^{-1})^{1/2}) \geq \lambda_j(A + 2A^{1/2} + I)$ .
Let $V_j$ be the span of the eigenvectors to the j largest eigenvalues of $A$ and $W_j = (I+B^{-1})^{-1/2}(V_j)$ .
Then
$\lambda_j((I+B^{-1})^{1/2}(A+B)(I+B^{-1})^{1/2})$ $ \geq min \{x^* ((I+B^{-1})^{1/2}(A+B)(I+B^{-1})^{1/2})x : x \in W_j$ and $x^* x = 1\}$ $\geq min \{x^* ((I+B^{-1})^{1/2}(\lambda_j(A)I+B)(I+B^{-1})^{1/2})x : x \in W_j$ and $x^* x = 1\}$ $\geq \lambda_j(A) + 2 \lambda_j(A)^{1/2} + 1 = \lambda_j(A + 2A^{1/2} + I)$ .
