Need some help with this one.
Given $A,E \in \mathbb{R}^{m\times n}$. Show that $$\sigma_{\max}(A+E) \leq \sigma_{\max}(A) + \|E\|_2 $$
The hint provided is: $$\sigma_{\min}(A)\|x\|_2 \leq \|Ax\|_2 \leq \sigma_{\max}(A)\|x\|_2$$
With the hint, I have tried \begin{align} \sigma_{\max}(A+E)\|x\|_2 &\leq \|(A+E)x\|_2 \\ &\leq \|Ax\|_2 + \|Ex\|_2 \\ \sigma_{\min}(A+E) &\leq \frac{\|Ax\|_2}{\||x\|_2} + \frac{\|Ex\|_2}{\|x\|_2} \\ &\leq \frac{\|Ax\|_2}{\|x\|_2} + \|E\|_2 \end{align}
Am I on the right track? How can I get $\sigma_{\max}(A)$ into this inequality?
You've already shown $$\frac{\|(A+E)x\|_2}{\|x\|_2} \le \frac{\|Ax\|_2}{\|x\|_2} + \frac{\|Ex\|_2}{\|x\|_2} \le \sigma_{\max}(A) + \sigma_{\max}(E)$$ for all nonzero $x$.
Note that when $x$ is the right singular vector of $A+E$ corresponding to its maximum singular value, then the left-hand side is $\sigma_{\max}(A+E)$.