Prove inequality [closed]

Question

I have to prove the following inequality (See Appendix A of Classical Fourier Analysis of Grafakos). $$ \left(\frac{(1+s/y)^y}{e^s}\right)^{2y} \leq (1+s)^2 / \exp(s), \quad s\geq 0 $$ and $$ \left(\frac{(1+s/y)^y}{e^s}\right)^{2y} \leq \exp(-s^2),\quad -y<s < 0. $$ How can I prove this?

Answer 1

The first inequality fails to hold e.g. for $s=4$ and $y=1/10$.

The second inequality can be rewritten as \begin{equation*} L(y)\le e^{-s^2} \tag{10}\label{10} \end{equation*} if $-y<s<0$, where \begin{equation*} L(y):=L(s,y):=\left(e^{-s} \left(\frac{s}{y}+1\right)^y\right)^{2 y}. \end{equation*} We have \begin{equation*} L_1(y):=\frac{L'(y)}{2 y}\,\left(e^{-s} \left(\frac{s+y}{y}\right)^y\right)^{-2 y} =2 \ln \left(\frac{s+y}{y}\right)-\frac{s (s+2 y)}{y (s+y)} \end{equation*} and \begin{equation*} L'_1(y)=\frac{s^3}{y^2 (s+y)^2}<0, \tag{20}\label{20} \end{equation*} since $y>-s$ and $s<0$, so that $L_1$ is decreasing (on $(-s,\infty)$). Also, $L_1(y)\to0$ as $y\to\infty$. So, $L_1>0$ and hence $L$ is increasing, to $e^{-s^2}$, as $y$ is increasing to $\infty$. So, \eqref{10} follows. This completes the proof of your second inequality. $\quad\Box$

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Your first inequality was actually stated only for $y\ge1$ in that book (and $s\ge0$).
Then inequality \eqref{20} will turn to its opposite, $L'_1(y)\ge0$. So, here $L_1$ is nondecreasing, and still $L_1(y)\to0$ as $y\to\infty$. So, $L_1<0$ and hence $L$ is decreasing. So, \begin{equation} L\le L(1)=e^{-2 s} (1 + s)^2\le (1+s)^2 / \exp(s). \quad\Box \end{equation}

Answer 2

For the second inequality, write $t:=-s$, then $y>t>0$ and we need to prove that $$ (1-t/y)^y \leq \exp{\left\{-\frac{t^2}{2y}-t\right\}}. $$ Taking logarithms of both sides and denoting $t/y=:\alpha \in (0, 1)$ we get an equivalent inequality $$ \log{(1-\alpha)}\leq -\alpha^2/2-\alpha, $$ which follows from the Taylor series expansion of the function $\log{(1-x)}$.