Let $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$, $v=\begin{bmatrix}x\\y\end{bmatrix}$. Show that $\|Av\|=M\|v\|$ for $M \in \mathbb{R}^+$.
Solution:\begin{align*}
\|Av\| &= \sqrt{\smash[b]{(ax+by)^2+(cx+dy)^2}} \le |ax+by| + |cx +dy|\\
&\le |a||x|+|b||y|+ |c||x|+|d||y| = (|a|+|c|)|x| + (|b|+|d|)|y|.
\end{align*}
Here, I do not understand how this inequality holds:$$
(|a|+|c|)|x| + (|b|+|d|)|y| \le (|a|+|b|+|c|+|d|)\sqrt{x^2 + y^2}.$$
Could someone elaborate on this?
Thank you in advance.
Note that \begin{equation} |x| \le \sqrt{x^2 + y^2}, \qquad |y| \le \sqrt{x^2 + y^2}. \end{equation} Then the inequality you wrote follows immediately.