Reverse Chernoff bound Is there an reverse Chernoff bound which bounds that the tail probability is at least so much.

i.e if $X_1,X_2,\ldots,X_n$ are independent binomial random variables and $\mu=\mathbb{E}[\sum_{i=1}^n X_i]$. Then can we prove $Pr[\sum_{i=1}^n X_i\geq (1+\delta)\mu]\geq f(\mu,\delta,n)$ for some function $f$.
Here is an explicit proof that a standard Chernoff bound is tight up to constant factors in the exponent for a particular range of the parameters. (In particular, whenever the variables are 0 or 1, and 1 with probability 1/2 or less, and $\epsilon\in(0,1/2)$, and the Chernoff upper bound is less than a constant.)

If you find a mistake, please let me know.

Lemma 1. (tightness of Chernoff bound)
Let $X$ be the average of $k$ independent, 0/1 random variables (r.v.).
For any $\epsilon\in(0,1/2]$ and $p\in(0,1/2]$,  assuming $\epsilon^2 p k \ge 3$,

(i)
If each r.v. is 1 with probability at most $p$, then
$$\displaystyle
\Pr[X\le (1-\epsilon)p]
~\ge~ 
\exp\big({-9\epsilon^2 pk}\big).$$

(ii)
If each r.v. is 1 with probability at least $p$, then
$$\displaystyle
\Pr[X\ge (1+\epsilon)p]
~\ge~ 
\exp\big({-9\epsilon^2 pk}\big).$$

Proof.
We use the following observation:

Claim 1. If $1\le \ell \le k-1$, then
$\displaystyle {k \choose \ell} ~\ge~ \frac{1}{e\sqrt{2\pi\ell}} \Big(\frac{k}{\ell}\Big)^{\ell} \Big(\frac{k}{k-\ell}\Big)^{k-\ell}$

Proof of Claim 1.
By Stirling's approximation, 
$i!=\sqrt{2\pi i}(i/e)^ie^\lambda$ where $\lambda\in[1/(12i+1),1/12i].$

Thus, $k\choose \ell$, which is  $\frac{k!}{\ell! (k-\ell)!}$, is at least
$$
\frac{\sqrt{2\pi k}\,(\frac{k}{e})^k}
{  \sqrt{2\pi \ell}\,(\frac{\ell}{e})^\ell
~~\sqrt{2\pi (k-\ell)}\,(\frac{k-\ell}{e})^{k-\ell} }
\exp\Big(\frac{1}{12k+1} - \frac{1}{12\ell} - \frac{1}{12(k-\ell)}\Big)$$
$$
~\ge~
\frac{1}{\sqrt{2\pi\ell}} \Big(\frac{k}{\ell}\Big)^{\ell} \Big(\frac{k}{k-\ell}\Big)^{k-\ell}e^{-1}.
$$
QED

Proof of Lemma 1 Part (i).
Without loss of generality assume each 0/1 random variable in the sum $X$
is 1 with probability exactly $p$.
Note $\Pr[X\le (1-\epsilon)p]$ equals the sum $\sum_{i = 0}^{\lfloor(1-\epsilon)pk\rfloor} \Pr[X=i/k]$,
and $\Pr[X=i/k] = {k \choose i} p^i (1-p)^{k-i}$.

Fix $\ell = \lfloor(1-2\epsilon)pk\rfloor+1$.
The terms in the sum are increasing,
so the terms with index $i\ge\ell$
each have value at least $\Pr[X=\ell/k]$,
so their sum has total value at least
$(\epsilon pk - 2) \Pr[X=\ell/k]$.
To complete the proof, we show that 
$$(\epsilon pk - 2) \Pr[X=\ell/k] ~\ge~ \exp({-9\epsilon^2 pk}).$$

The assumptions $\epsilon^2pk\ge 3$ and $\epsilon\le 1/2$
give $\epsilon pk \ge 6$,
so the left-hand side above is at least $\frac{2}{3}\epsilon pk\, {k \choose \ell} p^\ell(1-p)^{k-\ell}$.
Using Claim 1,
to bound $k\choose \ell$,
this is in turn at least $A\, B$
where
$A =  \frac{2}{3e}\epsilon p k/ \sqrt{2\pi \ell}$
and
$
B=
\big(\frac{k}{\ell}\big)^\ell 
\big(\frac{k}{k-\ell}\big)^{k-\ell}
p^\ell (1-p)^{k-\ell}.
$

To finish we show $A\ge \exp(-\epsilon^2pk)$ and $B \ge \exp(-8\epsilon^2 pk)$.

Claim 2.  $A \ge \exp({-\epsilon^2 pk})$

Proof of Claim 2.
The assumptions $\epsilon^2 pk \ge 3$ and $\epsilon\le 1/2$
imply
(i) $pk\ge 12$.

By definition, $\ell \le pk + 1$.
By (i), $p k \ge 12$.
Thus, (ii) $\ell \,\le\, 1.1 pk$.

Substituting the right-hand side of (ii) for $\ell$ in $A$ gives
(iii) $A \ge \frac{2}{3e} \epsilon \sqrt{p k / 2.2\pi}$.

The assumption, $\epsilon^2 pk \ge 3$,
implies $\epsilon\sqrt{ pk} \ge \sqrt 3$,
which with (iii) gives (iv) $A \ge \frac{2}{3e}\sqrt{3/2.2\pi} \ge 0.1$.

From $\epsilon^2pk \ge 3$ it follows that (v) $\exp(-\epsilon^2pk) \le \exp(-3) \le 0.04$.

(iv) and (v) together give the claim. QED

Claim 3.   $B\ge \exp({-8\epsilon^2 pk})$.

Proof of Claim 3.
Fix $\delta$ such that $\ell=(1-\delta)pk$.
The choice of $\ell$ implies $\delta\le 2\epsilon$,
so
the claim will hold as long as $B \ge \exp(-2\delta^2pk)$.
Taking each side of this latter inequality to the power $-1/\ell$ and simplifying,
it is equivalent to
$$
\frac{\ell}{p k}  \Big(\frac{k-\ell}{(1-p) k}\Big)^{k/\ell-1}
~\le~
\exp\Big(\frac{2\delta^2 pk}{\ell}\Big).
$$
Substituting $\ell= (1-\delta)pk$ and simplifying, it is equivalent to
$$
(1-\delta)  \Big(1+\frac{\delta p}{1-p}\Big)^{\displaystyle \frac{1}{(1-\delta)p}-1}
~\le~
\exp\Big(\frac{2\delta^2}{1-\delta}\Big).
$$
Taking the logarithm of both sides and using $\ln(1+z)\le z$ twice,
it will hold as long as
$$
-\delta\, +\,\frac{\delta p}{1-p}\Big(\frac{1}{(1-\delta)p}-1\Big)
~\le~
\frac{2\delta^2}{1-\delta}.
$$
The left-hand side above simplifies to $\delta^2/\,(1-p)(1-\delta)$,
which is less than $2\delta^2/(1-\delta)$ because $p\le 1/2$.
QED

Claims 2 and 3
imply $A B \ge \exp({-\epsilon^2pk})\exp({- 8\epsilon^2pk})$.
This implies part (i) of the lemma.

Proof of Lemma 1 Part (ii).
Without loss of generality assume each random variable is $1$ with probability exactly $p$.

Note $\Pr[X\ge (1+\epsilon)p] = \sum_{i = \lceil(1-\epsilon)pk\rceil}^n \Pr[X=i/k]$.
Fix $\hat\ell = \lceil (1+2\epsilon)pk \rceil - 1$.

The last $\epsilon pk$ terms in the sum
total at least  $(\epsilon pk-2)\Pr[X=\hat\ell/k]$,
which is at least $\exp({-9\epsilon^2 pk})$.
(The proof of that is the same as for (i), except with $\ell$ replaced by $\hat\ell$
and $\delta$ replaced by $-\hat\delta$ such that $\hat\ell = (1+\hat\delta)pk$.)
QED