Suppose $A$ is a $n \times n$ complex matrix and there exists a hermitian matrix X such that
$$\begin{pmatrix} I + X & A \\ A^* & I-X \end{pmatrix} \geq 0$$
Prove that for every $y \in \mathbb{C^n}$
$$|y^*Ay| \leq y^*y$$
This is a question from section 6 of the book "Matrix Theory" by Fuzhen Zhang. Here's what I've tried so far: Since $$\begin{pmatrix} I + X & A \\ A^* & I-X \end{pmatrix} \geq 0$$ we know there exists a contraction $C$ such that
$$A = (I + X)^{\frac{1}{2}}C(I - X)^{\frac{1}{2}}$$
Also X is hermitian and therefore has real eigenvalues. Let $1\geq \lambda_1 \geq \lambda_2 \geq ... \geq \lambda_n \geq -1$ be eigenvalues of $X$. It's known that there exists a unitary matrix $U$ such that
$$ D_1 = U^*(I + X)^{\frac{1}{2}}U = diag(\sqrt{1+\lambda_1}, \sqrt{1+\lambda_2}, ..., \sqrt{1+\lambda_n})$$ and $$ D_2 = U^*(I - X)^{\frac{1}{2}}U = diag(\sqrt{1-\lambda_1}, \sqrt{1-\lambda_2}, ..., \sqrt{1-\lambda_n})$$ Hence it suffices to show that for every $y \in \mathbb{C^n}$
$$|y^* D_1CD_2y| \leq y^*y$$ But I haven't been able to prove this. Additionally if $A$ is a contraction, the desired inequality holds but I haven't been able to prove that either.
For every vector $y$, pick a complex number $\omega$ on the unit circle such that $(y^\ast Ay)\,\omega=-|y^\ast Ay|$. Since the given block matrix is positive semidefinite, we have $$ \pmatrix{y^\ast &\omega^\ast y^\ast}\pmatrix{I+X&A\\ A^\ast &I-X}\pmatrix{y\\ \omega y}\ge0. $$ The LHS is equal to $2y^\ast y-2|y^\ast Ay|$. Now we are done.