I can't find a formal proof for this inequality:
$$ q \ln(1-q)- q \ln q - \ln (1-q) \leq \ln2 \\ 0\lt q \lt\ 1 $$
Using the exponentials I found this expression:
$$ \frac{(1-q)^{(q-1)}}{q^{q}} \leq 2 $$
but I think that this is not the right procedure for a right proof. Can anyone help me?
On
Often these problems are easier with an intuitive feel for why they might be true. In my case, that almost always boils down to a picture describing the problem. This isn't a proof in its own right, but it can help provide intuition for the proof's strategy. In my case, I started by graphing the difference between the left-hand-side and the right-hand-side as follows.
This numerical evidence strongly indicates that there is a unique local extremum at $q=\frac12$, and we might very well be able to use that in our proof. In practice, it turned out that the numerical evidence was correct, and a proof revolving around local extrema works just fine.
Our strategy then is to try to find the maxima (if they exist) for your function. If we define $$f(q)=q\ln(1-q)-q\ln(q)-\ln(1-q)-\ln(2)$$ then this amounts to finding where $f'(q)=0$, since $f$ is differentiable on the entire specified domain. If we find a unique local maximum and no minima, the local maximum must be a global maximum. We can readily compute that $$f'(q)=\ln\left(\frac{1-q}q\right),$$ and setting that equal to $0$ yields the unique solution $q=\frac12$. A little algebra allows us to conclude $f\left(\frac12\right)=0$, as desired. By checking the second derivative $$f''(q)=\frac1{q(q-1)}$$ we note that this is a local maximum rather than a local minimum. Since there are no other local extrema, $f(q)$ must be strictly less than $f\left(\frac12\right)$ for any $q\in(0,1)\setminus\{\frac12\}.$ Hence, $f(q)\leq0$ on the entire domain, as desired.
HINT
We have that
$$q \ln(1-q)- q \ln q - \ln (1-q) \leq \ln 2\iff -(1-q)\ln(1-q)-q\ln q\le \ln 2$$
and
$$f(x)=-(1-x)\ln (1-x)-x\ln x\implies f'(x)=\ln (1-x)-\ln x=0 \implies x=\frac12$$