Let $A, B\in \mathbb R^{m\times n}$ and $x\in \mathbb R^n$. Is it true that $$\Vert Ax\Vert_2\le \left(\Vert A-B\Vert_F+\Vert B\Vert_2\right)\Vert x\Vert_2?$$
where $\Vert B\Vert_2$ is a spectral norm, and $\Vert A-B\Vert_F$ is a Frobenius norm.
I know that $\Vert Ax \Vert_2\le \Vert A\Vert_F\Vert x\Vert_2$, and clearly $\Vert A\Vert_F=\Vert A-B+B\Vert_F\le \Vert A-B\Vert_F + \Vert B\Vert_F$. So $\Vert Ax\Vert_2\le (\Vert A-B\Vert_F + \Vert B\Vert_F)\Vert x\Vert_2.$ But $\Vert B\Vert_2 \le \Vert B\Vert_F$. So I'm not able to confirm the inequality above.
Is there an alternative way to prove the above inequality? Or can someone refute it? Thanks a lot!