Show probability bound. Let $Y$ be the number of successes in $n$ trials of a Bernoulli experiment with success probability $p$.  Show that:
$$ Pr(|\frac{Y}{n} - p |<e) \geq 1 - \frac{1}{4ne^2}$$
I tried starting with Chebychev and work backwards but I got stuck:
I used $np(1-p)$ for variance since binomial.
$$
\begin{align}
Pr(|Y-\mu|\geq a) &\leq \frac{np(1-p)}{a^2}\tag1\\
Pr(|\frac{Y}{n}-\frac{\mu}{n}|\geq \frac{a}{n}) &\leq \frac{np(1-p)}{a^2}\tag2\\
\end{align}
$$
Since for binomial $\mu/n=p$
$$
\begin{align}
Pr(|\frac{Y}{n}-p|\geq \frac{a}{n}) &\leq \frac{np(1-p)}{a^2}\tag3\\
1-Pr(|\frac{Y}{n}-p|\geq \frac{a}{n}) &\geq 1-\frac{np(1-p)}{a^2}\tag4\\
Pr(|\frac{Y}{n}-p|< \frac{a}{n}) &\geq 1-\frac{np(1-p)}{a^2}\tag5\\
\end{align}
$$
I then set $e=a/n$
$$
\begin{align}
Pr(|\frac{Y}{n}-p|< e) &\geq 1-\frac{np(1-p)}{(en)^2}\tag6\\
Pr(|\frac{Y}{n}-p|< e) &\geq 1-\frac{p(1-p)}{e^2n}\tag7\\
Pr(|\frac{Y}{n}-p|< e) &\geq \frac{e^2n-p(1-p)}{e^2n}\tag8\\
\end{align}
$$
I gave up here because I still need to flip the greater than in the prob and get the RHS to be exactly the same (how??) and I still don't really know if this is the right approach to start with anyway.

By the AM-GM inequality, $$\sqrt{p(1-p)} \le \frac{p+(1-p)}{2} = \frac12.$$
Therefore, $$p(1-p) \le \frac14.$$
\begin{align}
Pr(|\frac{Y}{n}-p| < e) &\geq 1-\frac{np(1-p)}{(en)^2}\\
&= 1-\frac{p(1-p)}{e^2n}\\
&\ge 1-\frac{1}{4e^2n}\\
\end{align}
