Let $\bf{A}$ be a semi-positive matrix, $\bf{I}$ is the identity matrix, and $\alpha \geq 0$. Let $\bf{x}$ and $\bf{y}$ be two vectors. I'm looking for a lower bound for the following quantity:
$${\bf Q} = ({\bf A} + \alpha {\bf I}) ({\bf x}-{\bf y}).$$
Can I say that the following holds true: $$\| {\bf Q} \| \geq \alpha \|{\bf x}-{\bf y} \|$$
EDIT: I managed to write the following as an attempt to prove the inequality.
Since ${\bf A} + \alpha \mathbf{I}$ a positive definite matrix, then its eigenvalue decomposition is given by \begin{align} {\bf A} + \alpha \mathbf{I} = \mathbf{P} \mathbf{\Sigma} \mathbf{P^T}, \end{align} where $\mathbf{P}$ is an orthogonal matrix and $\mathbf{\Sigma}$ is a diagonal matrix with the eigenvalues $\{\sigma_j\}_{j=1}^{d}$ on the diagonal. Note that $\mathbf{A}$ is a positive semi-definite matrix, then $\sigma_j \geq \alpha, ~\forall j \in [d]$. Let $\mathbf{y} = \mathbf{y}_1 - \mathbf{y}_2$, then we can write \begin{align}\label{f0} \nonumber &\|\mathbf{Q}\|\\ \nonumber &=\|\mathbf{P} \mathbf{\Sigma} \mathbf{P^T} \mathbf{y}\|\\ &=\|\mathbf{\Sigma} \mathbf{P^T} \mathbf{y}\|, \end{align} where we have used the fact that for an orthogonal matrix, we have $\|\mathbf{P} \mathbf{x}\| = \|\mathbf{x}\|$. Let $\mathbf{z} = \mathbf{P^T} \mathbf{y}$, then we can write \begin{align} \nonumber &\|\mathbf{Q} \|\\ \nonumber &=\|\mathbf{\Sigma} \mathbf{z}\|\\ \nonumber &= \sqrt{\sum_{j=1}^d \sigma_j^2 z_j^2}\\ &\geq \alpha \|\mathbf{z}\| = \alpha \|\mathbf{P^T} \mathbf{y}\| = \alpha \|\mathbf{y}_1-\mathbf{y}_2\| \end{align}
Does this make sense?