Show $||Ax+b|| < || A\left[x+ \mu D (Ax+b) \right]+b ||$, A is positive definite

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Given matrix $A \in \mathbb{R}^{M \times N}, D \in \mathbb{R}^{N \times M}$ , vector $x\in \mathbb{R}^N$ and $b \in \mathbb{R}^M$. What is condition of A which will lead to $$ ||Ax+b|| < || A\left[x+ \mu D(Ax+b) \right]+b || $$ and what is condition of A which will lead to $$ ||Ax+b|| > || A\left[x+ \mu D(Ax+b) \right]+b || $$ where $\mu$ is a very small positive constant.

Can we prove the inequality when $A$ is positive definite?

This problem is trivial when all entries are numbers: $$ ||ax+b|| < || a\left[x+ (\mu d)(ax+b) \right]+b || $$ if $a>0, d>0$.

In the case of matrix and vectors, one can use dot product $$ ||Ax+b||^2=Ax\cdot Ax + 2Ax\cdot b+ b\cdot b = x^T(A^T A)x + 2Ax\cdot b+ ||b||^2 $$ to analyze it, but no clue how to proceed further.