Estimation of a two-sided Chernoff bound I have come across this paper which uses the following version of the Chernoff bound (page 7):

If $X_{i}$ are i.i.d. with mean $\mu$, such that $X_{i}\in[0,1]$,
  then
$Pr\Bigg[\lnot\Bigg(\frac{\mu}{1+\delta}\leq\frac{1}{m}{\sum^m_{i=1}}X_{i}\leq
\mu(1+\delta)\Bigg)\Bigg]\leq
2e^{-\delta^{2}\mu m/12}$

Which is a two-sided multiplicative Chernoff bound. However it is not an ordinary two-sided bound in the sense that the left bound is not of the form $(1-\delta)\mu$ but $\frac{\mu}{1+\delta}$.
My question is: how did the author of this paper reach this version? Other bounds I've found are looser. Specifically in other textbooks instead of the number 12 in $2e^{-\delta^{2}\mu m/12}$ there are smaller numbers (which makes the bound looser).
The tightest bound I've found is for the one-sided Chernoff bound in the book Randomized Algorithms by Motwani and Raghavan (page 72):

$Pr[X>(1+\delta)\mu]<e^{−\delta^2\mu/4},0<\delta<2e−1$

Thank you! 
 A: You can get this bound using Hoeffding inequality (as it is mentioned in the paper) with some extra condition on the relation of $\mu$ and $\delta$. First see that:
$$
Pr\Bigg[\frac{1}{m}{\sum^m_{i=1}}X_{i}\geq
\mu(1+\delta)\Bigg]\leq \exp(-{2m\mu^2\delta^2})
$$
And for the other side we have:
$$
Pr\Bigg[\frac{\mu}{1+\delta}\geq\frac{1}{m}{\sum^m_{i=1}}X_{i}\Bigg]\leq
\exp(-\frac{2m\mu^2\delta^2}{(1+\delta)^2})
$$
Applying union bound we get:
$$
Pr\Bigg[\lnot\Bigg(\frac{\mu}{1+\delta}\leq\frac{1}{m}{\sum^m_{i=1}}X_{i}\leq
\mu(1+\delta)\Bigg)\Bigg]\leq \exp(-{2m\mu^2\delta^2})+\exp(-\frac{2m\mu^2\delta^2}{(1+\delta)^2}).
$$
Note that we  can show:
$$
\frac{2m\mu^2\delta^2}{(1+\delta)^2}<2m\mu^2\delta^2 
$$
and using that on:
$$
Pr\Bigg[\lnot\Bigg(\frac{\mu}{1+\delta}\leq\frac{1}{m}{\sum^m_{i=1}}X_{i}\leq
\mu(1+\delta)\Bigg)\Bigg]\leq 2\exp(-\frac{2m\mu^2\delta^2}{(1+\delta)^2}).
$$
The desired inequality follows if we prove the following
$$
\frac{2m\mu^2\delta^2}{(1+\delta)^2}>
\frac{\delta^{2}\mu m}{12}.
$$
But this is equivalent to:
$$
\frac{2\mu}{(1+\delta)^2}>
\frac{1}{12}\implies (1+\delta)^2<24\mu.
$$
Therefore under this extra condition $(1+\delta)^2<24\mu$, the desired inequality follows. 
