I came up with the following conjecture while tacking another problem:
Conjecture. Let $p \in [0, 1]$ and $q = 1-p$. Then
$$ (p \log p + q \log q)^2 \leq -\log(p^2 + q^2)\log 2 $$
A numerical experiment shows that this must be true, but I have no idea whether this can be proved with a clever idea or it requires some dirty work.
Remark 1. Jensen's inequality gives
$$ -\log(p^2 + q^2) \leq -(p \log p + q \log q) \leq \log 2, $$
telling that the conjecture is not quite trivial.
Remark 2. Writing $\mathrm{H}_{\alpha}(p) = \frac{1}{1-\alpha}\log(p^{\alpha}+q^{\alpha})$ for the binary Rényi entropy (with the cases $\alpha=0,1,\infty$ being defined via limit), the above inequality is equivalent to
$$ \mathrm{H}_1(p)^2 \leq \mathrm{H}_2(p)\mathrm{H}_0(p). $$
Numerical experiment suggests that $\alpha \mapsto \log \mathrm{H}_{\alpha}(p)$ is convex, from which the above inequality will follow. However, logarithmic Rényi entropy, $\log \mathrm{H}_{\alpha}(\mathcal{D})$ as a function of $\alpha$ for a general discrete distribution $\mathcal{D}$, is not necessarily concave, meaning that the above inequality is specific to binary case (i.e., $\mathcal{D}$ is Bernoulli).
Of course, it is possible that this inequality has nothing to do with entropy-based interpretation at all.
Remark 3. A slightly weaker inequality,
$$ \mathrm{H}_1(p) \leq \frac{\mathrm{H}_2(p) + \mathrm{H}_0(p)}{2},$$
can be proved directly by differentiation. (This is weaker because it can be proved assuming the conjecture and applying AM-GM inequality.)
Best Answer
Here is a ugly proof:
The non-trivial case is that $pq > 0$.
We use the result in this question. @Erik Satie's pointed out this result in here.
Fact 1: Let $x, y > 0$. Then $$x\ln x + y\ln y \ge (x + y)\ln(x + y - \sqrt{xy}).$$
By Fact 1, we have $$p \ln p + q \ln q \ge \ln(1 - \sqrt{pq}). \tag{1}$$
It suffices to prove that $$\ln^2(1 - \sqrt{pq}) \le - \ln(p^2 + q^2)\ln 2. \tag{2}$$
Let $u = \sqrt{pq}\in [0, 1/2]$. (2) is written as $$\ln^2(1 - u) \le -\ln(1 - 2u^2)\ln 2. \tag{3} $$
It suffices to prove that, for all $u\in [0, 1/2]$, $$-\ln (1 - u) \le \sqrt{-\ln(1 - 2u^2)\ln 2}.\tag{4}$$
Fact 2: For all $u\in [0, 1/2]$, $$\sqrt{-\ln(1 - 2u^2)\ln 2} \ge (8\ln 2 - 4)u^3 + (8 - 12\ln 2)u^2 + (6\ln 2 - 3)u.$$ (The proof is given at the end.)
By Fact 2, it suffices to prove that $$-\ln(1 - u) \le (8\ln 2 - 4)u^3 + (8 - 12\ln 2)u^2 + (6\ln 2 - 3)u. \tag{5}$$
Let $$f(u) := (8\ln 2 - 4)u^3 + (8 - 12\ln 2)u^2 + (6\ln 2 - 3)u + \ln (1 - u).$$
We have $$f'(u) = \frac{(1-2u)^2}{1-u}[6\ln 2 - 4 - (6\ln2 - 3)u].$$ Let $u_0 = \frac{6\ln 2 - 4}{6\ln 2 - 3}$. We have $f'(u_0) = 0$, and $f'(u) > 0$ on $[0, u_0)$, and $f'(u) < 0$ on $(u_0, 1/2)$. Also, $f(0) = f(1/2) = 0$. Thus, $f(u) \ge 0$ on $[0, 1/2]$.
We are done.
$\phantom{2}$
Proof of Fact 2:
It suffices to prove that $${-\ln(1 - 2u^2)\ln 2} \ge \left[(8\ln 2 - 4)u^3 + (8 - 12\ln 2)u^2 + (6\ln 2 - 3)u\right]^2.$$
Let $$F(u) := -\ln(1 - 2u^2)\ln 2 - \left[(8\ln 2 - 4)u^3 + (8 - 12\ln 2)u^2 + (6\ln 2 - 3)u\right]^2.$$
We have $$F'(u) = \frac{2u(1 - 2u)}{1 - 2u^2} G(u)$$ where $G(u)$ is a quintic polynomial.
We can prove that there exists $0 < u_1 < 1/2$ such that $G(u) > 0$ on $(0, u_1)$, and $G(u) < 0$ on $(u_1, 1/2)$. (Note: We omit the proof here. )
Thus, we have $F'(u) > 0$ on $(0, u_1)$, and $F'(u) < 0$ on $(u_1, 1/2)$. Also, $F(0) = F(1/2) = 0$. Thus, $F(u) \ge 0$ on $[0, 1/2]$.
We are done.