Consider $$ \begin{gathered} \dot{x}_1=x_1 x_2 \\ \dot{x}_2=-x_2 . \end{gathered} $$
Show that $$ \left|x_1(t)\right|+\left|x_2(t)\right| \leq \alpha\left(\left|x_1(0)\right|+\left|x_2(0)\right|\right), $$ where $$ \alpha(s) \triangleq s e^s . $$
Here $\alpha$ denotes K class function.
I found solution of $x_2$ and have $x_2(t)=x_2(0)e^{-t}$. Then I plugged that in the first equation and got $x_1(t)=x_1(0)e^{e^{-t}}$. I am not sure how to proceed with this proof.
First, a correction: the solution to $\dot{x}_1=x_1x_2=x_1x_2(0)e^{-t}$ is $$ x_1(t)=x_1(0)e^{x_2(0)(1-e^{-t})}. \tag{1} $$ Now we have a sequence of straightforward inequalities: if $t\geq 0$, then $$ |x_2(t)|=|x_2(0)e^{-t}|\leq |x_2(0)| \tag{2} $$ and $$ |x_1(t)|=\left|x_1(0)e^{x_2(0)(1-e^{-t})} \right| \leq |x_1(0)|e^{|x_2(0)(1-e^{-t})|} \leq |x_1(0)|e^{|x_2(0)|}. \tag{3} $$ Therefore, \begin{align} |x_1(t)|+|x_2(t)|&\leq |x_1(0)|e^{|x_2(0)|}+|x_2(0)| \\ &\leq (|x_1(0)|+|x_2(0)|)e^{|x_2(0)|} \\ &\leq (|x_1(0)|+|x_2(0)|)e^{|x_1(0)|+|x_2(0)|} \\ &=\alpha(|x_1(0)|+|x_2(0)|). \quad{\square} \tag{4} \end{align}