determinant inequality for Hermitian matrix $A \in \mathbb{C}^{M \times M}$ is a positive semidefinite matrix with all diagonal entries being $1$. and the vector $\mathbf{y} \in \mathbb{C}^{M}$ has entries $|y_{i}| < 1$. Prove that
$$2 \det(A) - \det(A + yy^H) \geq 0$$
I tried to use interlacing theorem but got stuck.
Please help me.

I don't think this is true. Let $$ A=\begin{bmatrix}1&1\\1&1\end{bmatrix}, \ \ \ y=(1,0)^T.$$
Then
$$
yy^T=\begin{bmatrix}1&0\\0&0\end{bmatrix},\ \ \ A+yy^T=\begin{bmatrix}2&1\\1&1\end{bmatrix}.
$$
Then
$$
2\det A-\det(A+yy^T)=0-1=-1.
$$
