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 $x^* x = 1\}$
$\geq \lambda_j(A) + 2 \lambda_j(A)^{1/2} + 1 = \lambda_j(A + 2A^{1/2} + I)$ .