For two complex matrices $A,B \in \mathbb{C}^{n\times m}$ how to prove that:
\begin{equation}
\overline{\sigma}(B-A) \ge \underline{\sigma}(B) - \underline{\sigma}(A) 
\end{equation}
where $\underline{\sigma}(A)$ and $\overline{\sigma}(A)$ denote the smallest and respectively the largest [singular value](https://en.wikipedia.org/wiki/Singular_value) of $A$.



Inequality between the singular values for a sum of two matrices
We have $\underline{\sigma}(C)=\min\_{x:\|x\|=1} \|Cx\|$, $\overline{\sigma}(C)=\max\_{x:\|x\|=1} \|Cx\|$. Now the inequality is a partial case of a general inequality $\max (f-g)\geqslant \min(f)-\min(g)$ for any two real-valued functions $f,g$ on any set. This is seen from writing $g(x)=f(x)-(f-g)(x) \geqslant \min f-\max (f-g)$ for a fixed $x$ and then taking minimum over $x$. 


