Let $\mathcal{A} = \{0, 1\}$ consider two distributions $P$ and $Q$ on $\mathcal{A}$. Let $P(0) = p$ and $Q(0) = q$. Please, prove that $$D(P||Q) > 2(p - q)^2 \textbf{ log}e.$$
I started with $D(P||Q) = P(0)$log$\frac{P(0)}{Q(0)} + P(1)$log$\frac{P(1)}{Q(1)}$
$= p$log$\frac{p}{q} + P(1)$log$\frac{P(1)}{Q(1)} = p$log$\frac{p}{q} + (1 - p)$log$\frac{1 - p}{1 - q}$
$= p$log$(p) -p$log$(q) + (1 - p)$log$(1 - p) -(1 - p)$log$(1 - q)$
$= p$log$(p) -p$log$(q) + $log$(1 - p) -$log$(1 - q) -p$log$(1 - p) + p$log$(1 - q)$
Now since I know that log is a concave function I was thinking rewrite
$p$log$(p) -p$log$(q) + $log$(1 - p) -$log$(1 - q) -p$log$(1 - p) + p$log$(1 - q)$
in the form of
log$((1 - \alpha)x +\alpha y)$
and re-write
$2(p - q)^2$log$e$
in the from
$(1-\alpha)$log$ x + $ log$y$
for some $\alpha \in [0,1]$. I was also thinking since I know $p \in [0,1]$ that $p$ would be my most logical choice for $\alpha$ and then if I set $x = \frac{1 - p}{1 - q}$ and set $y = \frac{p}{q}$ then I would have my left had side in the form I want the right hand side to be on, and I still don't know what to do with $2(p - q)^2$log$e$.
Edit: So I decided to try to let $f(p,q) = p$log$\frac{p}{q} + (1- p)$log$(\frac{1-p}{1-q}) - 2(p - q)^2$log$e$
Then $f_p =$ log$\frac{p}{q} + p(\frac{q}{p})\frac{1}{q} -$log$(\frac{1-p}{1-q}) + (1- p)\frac{1 - q}{1 -p}(\frac{-1}{1-q}) - 4(p-q)$log$e$
$=$ log$\frac{p}{q} + 1 -$log$(\frac{1-p}{1-q}) -1 - 4(p-q)$log$e$. And
$f_{pp} = \frac{q}{p}(\frac{1}{q}) - \frac{1- q}{1-p}(\frac{-1}{1-q}) - 4$log$e = \frac{1}{p} + \frac{1}{1- p} - 4$log$e = \frac{1}{p(1-p)} -4$log$e$.
Also $f_q = p\frac{q}{p}(\frac{-p}{q^2}) +(1-p)\frac{1-q}{1-p}(\frac{1 - p}{(1-q)^2} + 4(p - q)$log$e = -\frac{p}{q} + \frac{1 - p}{1-q} + 4(p-q)$log$e$
And $f_{qq} = \frac{p}{q^2} + \frac{1-p}{(1- q)^2} - 4$log$e$.
So $0 = f_{pp}$ iff $0 = 1 - 4p(1-p)$log$e = 4p^2$log$e -4p$log$e + 1$. So $p = \frac{4log e \pm \sqrt{16log^2e - 16log e}}{8loge}$
But I'm still not sure how this helps me
This is one of those situations where it's easy to get bogged down in pages of calculations without any progress. Your observation of $\log$ being concave and the strategy is a good idea. However, one can go even simpler, by essentially viewing the two-dimensional problem ($p,q$) in sections - i.e., fixing a $p$ and then studying the one-dimensional problem on only $q$ (this doesn't always work out, of course, but in this case it does, and is usually much simpler than dealing with the full problem, so is worth trying). After adopting this strategy it's just a matter of making some simplifying notation changes and using your intuition from single variable calculus.
Firstly, note that we only need to show this for one base of the logarithm, since for other bases the result follows by the change of base relation for $\log$s. So lets fix the natural base (for easy derivatives).
Let's fix a $p$, and study how $D$ changes as $q$ varies. In particular, setting $q = p+\varepsilon,$ consider $$ f_p(\varepsilon) := D(p\|p+\varepsilon) - 2\varepsilon^2 = -H(p) - p\log (p + \varepsilon) - (1-p) \log (1 - p - \varepsilon) - 2\varepsilon^2, $$ where $\varepsilon \in (-p, 1-p)$.
We want to show that for all valid $\varepsilon, p$, $f_p \ge 0$. Notice that $f_p(0) = 0$ for every $p$. Further, $$ \partial_{\varepsilon} f_p(\varepsilon) = \frac{1-p}{1-p - \varepsilon} -\frac{p}{p+\varepsilon} - 4\varepsilon \\= \varepsilon \left( \frac{1}{(p+\varepsilon)(1-p-\varepsilon)} - 4\right) =: \varepsilon g_p(\varepsilon).$$
Suppose we can show that $g_p(\varepsilon) \ge 0$. Then we'll be done, because this means that $f_p$ is decreasing for negative $\varepsilon,$ and increasing for positive $\varepsilon,$ which means its minimum is at $\varepsilon = 0,$ which we've already seen to be $0$.
But since $\varepsilon \in (-p, 1-p)$, both $p+\varepsilon$ and $1- p - \varepsilon$ are non-negative. So, using the AM-GM inequality, we find that $$ \sqrt{(p+\varepsilon)(1-p-\varepsilon)} \le \frac{1 - p - \varepsilon + p + \varepsilon}{2} = \frac{1}{2},$$ which can be directly rearranged into $g_p(\varepsilon) \ge 0,$ concluding the argument.