Let $X : \Omega \to \{0,1\}$ be a Bernoulli random variable wich takes value $0$ with probability $p$ and value $1$ with probability $q$, $p+q=1, p,q \geq 0$. Your inequality is the claim that
$$
\mathbb{E} f^{2}(X) \ln f^{2}(x) - \mathbb{E}f^{2}(X) \ln \mathbb{E} f^{2}(X) \leq pq \frac{\ln p - \ln q}{p -q}\mathbb{E} |Df(X)|^{2}
$$
for all $f : \{-1,1\} \to \mathbb{R}$ where in this simple case
$$
|Df(x)|^{2}=|f(1)-f(0)|^{2}.
$$
For the proof of this and its consequences (Gaussian measure $p=q=2$, Poisson measure arbitrary $p,q$) see the paper of S. Bobkov and M. Ledoux, On Modified Logarithmic Sobolev Inequalities for Bernoulli and Poisson Measures
These kind of two-point inequalities are quite subtle and sometimes hard to prove, but if applied properly they can give interesting and unexpected applications for Gaussian inequalities and not only.
There is one similar two-point inequality related to complex hypercontractivity on the discrete cube which is still open and if somebody will be interested I can mention it.
UPDATE:
Let $1< p \leq q < \infty$, and $|z|\leq 1, z \in \mathbb{C}$. The following conditions are equivalent:
(i) For all $a,b \in \mathbb{C}$ we have $$ \left( \frac{|a+bz|^{q}+|a-bz|^{q}}{2} \right)^{1/q} \leq \left(\frac{|a+b|^{p}+|a-b|^{p}}{2} \right)^{1/p} $$
(ii) for all $w \in \mathbb{C}$ we have $$ (q-2)(\Re\; wz)^{2} + |wz|^{2} \leq (p-2)(\Re w)^{2}+ |w|^{2} $$
Remark: (i) $\Rightarrow$ (ii), follows by Taylor's formula (take $a=1$ and $b \to 0$). The implication (i) $\Rightarrow$ (ii) is open for $3/2 <p\leq q<2$ (and its dual part $2<p\leq q <3$). For the remaining part of exponents this is the result of F. Weissler Two-point inequalities, the Hermite semigroup and the Gauss-Weierstrass semigroup
By the way, the particular case $z=i\sqrt{p-1}$, $1<p\leq 2$, and $q=p/(p-1)$ gives (after proper application of Minkowski's inequality and CLT) Hausdorff--Young inequality with sharp constants proved by W. Beckner.