Lower bound on norm of product of two matrices

The inequality is true. It is obvious when $A$ is singular. When $A$ is invertible, for any unit vector $x$, we have $\|x^TA\|\ge\sigma_\min(A)$. Therefore \begin{align} \|AB\| &= \max_{\|x\|=1} \|x^TAB\|\\ &= \max_{\|x\|=1} \|x^TA\|\left\|\frac{x^TA}{\|x^TA\|}B\right\|\tag{1}\\ &\ge \max_{\|x\|=1} \sigma_\min(A)\left\|\frac{x^TA}{\|x^TA\|}B\right\|\\ &= \max_{\|y\|=1} \sigma_\min(A)\left\|y^TB\right\|\tag{2}\\ &= \sigma_\min(A)\|B\|, \end{align} where we have used the assumption that $A$ is invertible in $(1)$ (so that we can divide by $\|x^TA\|\neq0$) and $(2)$ (so that every $x$ corresponds to a unique $y$ and vice versa).

By use of the multiplicativity of the spectral norm: $$\|ABB^{-1}\|_2 \le \|AB\|_2\|B^{-1}\|_2,$$ which implies that $$\|AB\|_2\ge \|A\|_2 \frac{1}{\|B^{-1}\|_2}.$$