Elementary proof of $\left(1+\frac{t}{\sqrt{n}}\right)^{n}e^{-\sqrt{n}t}\le\left(1+t\right)e^{-t}$ for $n\in\mathbb{N},t\ge0$ .

Question

I am looking for an elementary proof for the inequality

$\left(1+\frac{t}{\sqrt{n}}\right)^{n}e^{-\sqrt{n}t}\leq\left(1+t\right)e^{-t}$ for $n\in\mathbb{N},t\geq0 $

I encountered this inequality while writing an advanced exercise for first year students. I assumed it can be done by defining the function

$$\varphi\left(x\right)=\left(1+\frac{t}{x}\right)^{x^{2}}e^{-xt}\,,$$

and show it monotonically decreases for $x\geq1$ and positive $t$, but failed to do so. Does anyone know an elementary way to prove the above inequality?

Answer 1

Not sure what you mean by elementary, but if $$f(t)=\left(1+\frac{t}{\sqrt{n}}\right)^n e^{-\sqrt{n}t} \\ g(t)=(1+t)e^{-t}\, ,$$ then you want $\log (f) \leq \log (g)$, or $$0\leq \log(1+t)-t-n\log\left(1+\frac{t}{\sqrt{n}}\right)+\sqrt{n}t = h(t) \, .$$ We have $h(0)=0$ and $$h'(t)=\frac{t^2 (\sqrt{n}-1)}{(1+t)(\sqrt{n}+t)} \geq 0 \, .$$

Answer 2

Following your original idea, let consider for $x\ge 0$

$$f(x)=\left(1+\frac1x\right)^{x^2}e^{-x} \implies f'(x)=\frac{f(x)}{2x(x+1)^2}\left( \log\left( 1+\frac1x\right) -\frac{2x+1}{2x(x+1)}\right)\le 0$$

indeed, following the suggestion given by Diger, we have

$$g(x)=\log \left(1+\frac1x\right)-\frac{2x+1}{2x(x+1)}\implies g'(x)=\frac{1}{2x^2(x+1)^2}\geq 0$$

with $ \lim_{x\to 0^+}g(x) = -\infty$ and $\lim_{x\to\infty}g(x) = 0$.

Hence $f(x)$ is decreasing and therefore by $x=\frac{\sqrt{n}}t$ also the sequence

$$b_n=\left(1+\frac{t}{\sqrt{n}}\right)^{\frac n {t^2}}e^{-\frac{\sqrt{n}}t}$$

is decreasing and then also

$$a_n=(b_n)^{t^2}=\left(1+\frac{t}{\sqrt{n}}\right)^{n}e^{-\sqrt{n}t}$$

is decreasing, therefore $a_n \le a_1$.