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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!

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    $\begingroup$ for $t=\frac{y}{1+x}$, $f = y^{-1} t e^(-t)$. but since$xe^(-x)$ has its maximum on 1... $\endgroup$ Jan 31, 2020 at 1:11

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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} $.

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