On Chernoff bound Suppose $X\sim \textrm{Binomal}(n,p)$, show that for $\epsilon\geq 1$
$$P\bigg\{{X/n \over p}\geq \epsilon\bigg\}\leq \exp\bigg[-np(\epsilon(\log \epsilon -1)+1)\bigg]$$
I suppose there is a way of using Chernoff Bound of some kind but I don't quite get the right procedure. Any hint or comment will be appreciated.
 A: The Chernoff inequality asserts
$$
    \Pr(X > a) \leqslant \min_{t > 0} \mathsf{E}\left(\mathrm{e}^{t (X-a)}\right) \tag{1}
$$
Since the moment generating function of the binomial distribution is $$\mathrm{E}(\exp(t X)) = \left(1 + \left(\mathrm{e}^t-1\right)p \right)^n$$
The left-hand-side of the inequality $(1)$ for $a = n p \epsilon$ reads:
$$
    \min_{t > 0} \left( \mathrm{e}^{-t p \epsilon} \left(1 -p + p \mathrm{e}^{t}\right) \right)^n =  \left( \min_{t > 0} \mathrm{e}^{-t p \epsilon} \left(1 -p + p \mathrm{e}^{t}\right) \right)^n
$$
Differentiating the objective function with respect to $t$ and solving the equation for extremum gives:
$$
    \exp(t_\max) = \epsilon \frac{1-p}{1-p \epsilon}
$$
Restriction of $t_\max>0$ imposes $\epsilon > 1$. This gives:
$$
   \Pr\left(X > \epsilon n p\right) \leqslant \left(\left(\epsilon \frac{1-p}{1-p \epsilon}\right)^{-p \epsilon} \frac{1-p}{1-p \epsilon}\right)^n = \mathrm{e}^{-n p \epsilon \log \epsilon} \left(\frac{1-p}{1-p \epsilon}\right)^{n (1-p \epsilon )} \tag{2}
$$
Using $$ \left(\frac{1-p}{1-p \epsilon}\right)^{n(1-p \epsilon)} = \left( 1 + \frac{n p(\epsilon-1)}{n (1-p \epsilon) } \right)^{n(1-p \epsilon)} \leqslant \mathrm{e}^{n p (\epsilon-1)}  \tag{3} $$
where the last inequality follows from $\forall_{a\geqslant 0,x>0} \left(1+\frac{a}{x}\right)^x \leqslant \mathrm{e}^{a}$. 
Combining eqs. $(2)$ and $(3)$ we get the desired inequality:
$$
   \Pr(X > \epsilon n p) \leqslant \exp\left(- n p \left( \epsilon \log \epsilon + 1 - \epsilon \right) \right)
$$
