# Is there any intuitive way to see $\frac{e^{\delta}}{(1+\delta)^{1+\delta}}\leq e^{-\delta^2/3}$ $0<\delta<1$

I am reading the proof of Chernoff bound ,there's one step here : $$\frac{e^{\delta}}{(1+\delta)^{1+\delta}}\leq e^{-\delta^2/3}$$ where $0<\delta<1$

the book prove that by using the second order dervative analysis of

$$f(\delta)=\delta -(1+\delta)log(1+\delta)+\frac{\delta^2}{3}$$

but how to see this upper bound for$\frac{e^{\delta}}{(1+\delta)^{1+\delta}}$ or is there any straight inequality manipulation to get there?

• The inequality is false for $\delta=2$. Do you have a restricted range of $\delta$ in mind? – Greg Martin Sep 8 '18 at 0:45
• @GregMartin thank you are right , I already fixed it – ShaoyuPei Sep 8 '18 at 0:47

Instead of comparing $$\frac{e^{\delta}}{(1+\delta)^{1+\delta}}\leq e^{-\delta^2/3}$$ take logarithms and prove that $$\delta -(1+\delta ) \log (1+\delta ) < -\frac{\delta^2}3$$ Expanded as series built around $\delta=0$, the lhs is $$-\frac{\delta ^2}{2}+\frac{\delta ^3}{6}-\frac{\delta ^4}{12}+O\left(\delta ^5\right)$$ Now, use the fact that $\delta<1$ making $$-\frac{\delta ^2}{2}+\frac{\delta ^3}{6} \le -\frac{\delta ^2}{2}+\frac{\delta ^2}{6}=-\frac{\delta ^2}{3}$$