Tight upper bound for the binary entropy function The binary entropy function is defined as $H_2(x) = -x\log_2(x) - (1-x)\log_2(1-x)$. I am aware of the upper bound $2\sqrt{x(1-x)}$ provided in this question already.
One of the answers implies that $H_2(x) \leq (4x(1-x))^{\frac{3}{4}} + 0.008$. From Wolfram Alpha here, it seems obvious that in fact $H_2(x) \leq (4x(1-x))^{\frac{3}{4}} + 0.007$. Is there a nice way to show this?
 A: There is a more systematic treatment of this question which I can refer to here. So this "replaces" in a way the precise bound in the question, I hope this is fine with OP.
One can ask the following question: suppose we want to establish an upper bound
$$H_2(x) \leq c \cdot (x(1-x))^{q} \quad ,$$
with some constant $c$, what is the largest $q^*$ where this is possible?
You have already noticed that $q= 0.5$ is possible. You tried to improve that and found that  $q= 0.75$ is "almost" possible. To make it strict,  you (and Jack D'Aurizio) needed an extra additive constant.
The following results are taken from the paper  "Bounds for entropy and divergence for distributions over a two-element set" by Topsøe, Flemming. In:  JIPAM. Journal of Inequalities in Pure & Applied Mathematics. 2 (2): Paper No. 25 (2001). Link to pdf. Note when you read the paper that Topsøe uses natural logarithms so a little transformation is necessary.
From Topsøe's paper, the answer to the above question is $q^* = 1 / \ln(4) \simeq 0.721$ (which is close to your $0.75$). In this case, $c^* =  4^{q^*} = e \simeq 2.718 $, which can be compared to the constant in your proposition, $c = 4^{3/4} \simeq 2.82$, again very close.
Since $c^* =  4^{q^*}$ this is also a very good reason  to include a factor $4$ into the RHS of the proposition, which you and Topsøe have done. Another reason is establishing the equality  at $x^* = 0.5$, where for all $q$ holds:
$1 = H_2(x^*) =  (4 \, x^*(1-x^*))^{q} $.
Again, I modified your question for the reason of improving the somewhat arbitrary constant + exponent proposition and to make the treatment more concise, hope that is fine with you.
