I have a two-variable function $f(x,y)$ defined as:
\begin{equation}
f(x,y) := \frac{1}{x+1}\,\exp\left(-\frac{y}{x+1}\right), x\geq 0,\, y\geq 0.
\end{equation}
I want to show that this function for all $y\geq 1$ is upper bounded by $\left(y\exp(1)\right)^{-1}$. However, I am not able to find a starting point. Any help would be highly appreciated!
Best Answer
If $g(x) =xe^{y/x} $ (essentially the reciprocal of this, , then $g'(x) =\dfrac{e^{y/x} (x - y)}{x} $.
$g'(x) = 0$ at $x=y$ and is $> 0$ for $x > y$ and $< 0$ for $x < y$. Therefore $g(x)$ has a minimum at $x = y$ of $ey$, so $\dfrac1{g(x)} \le \dfrac1{ey} $.