Why $(1 + \epsilon)^{k/2} \exp (-\epsilon)^{k/2} < \exp\left(-\frac{k}{2}(\epsilon^2/2-\epsilon^3/3)\right)$ I am reading that given $\epsilon > 0$ and $k> 0$, by series expansions we get
$$(1 + \epsilon)^{k/2} \exp (-\epsilon)^{k/2} < \exp\left(-\frac{k}{2}(\epsilon^2/2-\epsilon^3/3)\right).$$
Could you please someone explain how we can prove that? Any help is highly appreciated!
 A: You can rewrite the inequality :
$$ (1+x)^{k/2}<\exp(\dfrac{k}{2}(\epsilon-\epsilon^2/2+\epsilon^3/3))$$

*

*You an write $(1+\epsilon)^{k/2}=\exp(\dfrac{k}{2}(\ln(1+\epsilon))$

*Consider the Taylor-Lagrange inequality between $(1+x)$ and $1$ at second maximum order so.

$$ |\ln(1+x)-\ln(1)-\sum_{k=1}^n\dfrac{x^k}{n!}\ln^{(k)}(0) |\leq \mathcal{M}\dfrac{x^{n+1}}{(n+1)!}$$
For $n=2$, $\mathcal(M)$ being the sup of $x \to \ln^{(n+1)}(x)$
$$ |\ln(1+x)-x+\dfrac{x^2}{2} |\leq \dfrac{2x^{3}}{(3)!}$$
Getting the double inequality of absolute value :
$$ x-\dfrac{x^2}{2}-\dfrac{x^{3}}{3}<\ln(1+x)\leq x-\dfrac{x^2}{2}+\dfrac{x^{3}}{3}$$
Which end the problem.
A: Since $\ (\epsilon^2 - \epsilon + 1) ( 1 + \epsilon) =  \epsilon^3 + 1 > 1\quad \forall\ \epsilon > 0,\ $ it follows that:
$$ \epsilon^2-\epsilon + 1 > \frac{1}{1 + \varepsilon} \quad \forall\ \epsilon > 0. $$
So, if we define $\ f(\epsilon):= \frac{\epsilon^3}{3} - \frac{\epsilon^2}{2} + \epsilon,\ $ and $\ g(\epsilon) := \ln(1+\epsilon),\ $ then the above tells us that:
$$\ f'(\epsilon) > g'(\epsilon)\ \text{ i.e. }\ f'(\epsilon) - g'(\epsilon)> 0\quad \forall\ \epsilon > 0.$$
Note also that $\ f(0) = g(0),\ $ i.e. $\ f(0) - g(0) = 0.\ $
Define $\ h(\epsilon):= f(\epsilon) - g(\epsilon).\ $ Then $\ h(0) = 0\ $ and $\ h'(\epsilon) > 0\ \forall\ \epsilon>0.\ $ Now suppose $\ \exists\ c>0\ $ such that $\ h(c)\leq 0.\ $ Then by MVT on $\ [0,c],\ \exists\ c' \in (0,c)\ $ such that $\ h'(c) \leq 0,\ $ a contradiction. This shows that $\ h(\epsilon) > 0\ \forall \epsilon>0,\ $ i.e. $\ f(\epsilon) > g(\epsilon)\ \forall\ \epsilon>0,\ $ i.e. :
$$ \frac{\epsilon^3}{3} - \frac{\epsilon^2}{2} + \epsilon > \ln(1+\epsilon)\quad \forall\ \epsilon > 0.\qquad (1) $$
Multiplying both sides of $\ (1)\ $ by $\ \frac{-k\epsilon  }{2}\ $ and rearranging, we get:
$$ \frac{k}{2} \ln(1+\epsilon) + \left(-\epsilon\frac{k}{2}\right) < -\frac{k}{2}\left( \frac{\epsilon^2}{2} - \frac{\epsilon^3}{3} \right). $$
Due to $\ g(x) = e^x\ $ being an increasing function, we may exponentiate both sides, and this yields the result.
