Show $|\exp(-x/2) - \exp(-y/2)| \leq |x-y|/4$ for $x,y\geq 0$. I am trying to show this inequality:
$$
\left|e^{-x/2} - e^{-y/2}\right| \leq \frac{|x-y|}{4}
$$
for $x,y\geq 0$. 
I've gotten stuck and could use some kind assistance. 
Many thanks in advanced!

This is incorrect as written. E.g., for $x = 0, y = 1$,  $$LHS = 1 - e^{-1/2} \approx 0.393 \color{red}{>} \frac{1}{4} = RHS$$
However it is true that 
$$\left|e^{-x/2} - e^{-y/2}\right| \leq \frac{|x-y|}{2} \quad\text{ for all } x, y \geq 0$$
To see this, write $f(x) = e^{-x/2}$. Then by the Mean Value Theorem, for all $0 \leq x < y$,
$$\frac{e^{-x/2} - e^{-y/2}}{x-y} = f'(c) \quad\text{ for some } c \in (x,y)$$ The absolute value of the derivative $\left|f'\right|$ is bounded above on the non-negative reals by $1/2$. Hence
$$\left|\frac{e^{-x/2} - e^{-y/2}}{x-y}\right| \leq \frac 1 2$$

One other alternative: if you require the bound to be
$$\left|e^{-x/2} - e^{-y/2}\right| \leq \frac{|x-y|}{4}$$
 then this will the case for $x, y \geq a$ such that $\left|f'(a)\right| = \frac 14$, i.e., $x, y \geq 2 \ln 2$.
