an elementary inequality in two variables (absolute values and exponentials)

Question

I'm interested in approaches to proving the inequality $$ |1-xe^y| \le (|1-x| + |y|) \, e^{|y|} \qquad (x,y \text{ real}). $$ At first glance it looks like it might follow almost immediately from the triangle inequality but I do not see the details. Should it be broken into cases? (When $y>0$ and $x>1$ it reduces to the easy relation $-1 \le (y-1) e^y$, but some other cases do not appear to be as simple.)

Answer 1

Since $|1-xe^y|=|1-e^y+e^y-xe^y|\leq|1-e^y|+|1-x|e^y\leq|1-e^y|+|1-x|e^{|y|}$, it is enough to prove $|1-e^y|\leq|y|e^{|y|}$.

If $y\geq0$, we have $-1\leq(y-1)e^y$, which holds because $(y-1)e^y$ has its minimum $-1$ at $y=0$.

If $y<0$, we have $1\leq(-y+1)e^{-y}$, which holds because the derivative of the right side is $(y-2)e^{-y}<0$ and equality holds when $y=0$.