Exercise related to log-Sobolev inequalities This is essentially what Exercise 5.4 in 
Boucheron, Lugosi, Massart Concentration Inequalities boils down to:
For real $a,b$ and $0<p<1$,
\begin{align*}
&pa^2\log( \frac{a^2}{b^2+pa^2-pb^2}) + 
(1-p)b^2\log(\frac{b^2}{b^2+pa^2-pb^2})
\\
&\le
\frac{p(1-p)(a-b)^2}{1-2p}\log\frac{1-p}p 
.
\end{align*}
This is supposed to be provable by elementary means and in finite time. Any ideas?
 A: 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
$$
\mathbb{E} f^{2}(X) \ln f^{2}(x) - \mathbb{E}f^{2}(X) \ln \mathbb{E} f^{2}(X) \leq  \frac{\ln p - \ln q}{p -q}\mathbb{E} |Df(X)|^{2}
$$
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.
A: After Gerhard Paseman's and Nate Eldredge's suggestions, the problem is reduced to showing that
$$ f_{a,b}(p)=
%A^2 p Log[A^2/1]
p a^2 \log(a^2)
+
%
(1-p) b^2\log(b^2)
-
\frac{p(1-p)(a-b)^2}{1-2p}\log\frac{1-p}p 
\le0 
$$
for $a,b>0$ and $0<p<1/2$.
The claim is obviously true for $p\to0$ and less obviously true for $p\to1/2$
(this is the content of Theorem 5.1 in the Boucheron, Lugosi, Massart book). Furthermore, a straightforward calculation yields
$$
f''_{a,b}(p) =
\frac{
(a - b)^2 (1 - 2 p - 2 (1 - p) p 
\log[(1-p)/p)]
}{
(1 - p) p (1 -
   2 p)^3
},
$$
which we claim is nonnegative for $0<p<1/2$. Once the latter claim is established, we have that $f_{a,b}(p)\le0$ for $p=0,1/2$ and is convex on $[0,1/2]$, so it is nonpositive on the whole interval.
Proving the nonnegativity of the second derivative is quite straightforward, if somewhat tedious. The denominator in the expression for $f''_{a,b}$ is obviously positive, so it suffices to consider just the numerator (and disregard the $(a-b)^2$ factor).
The expression in question, $1 - 2 p - 2 (1 - p) p 
\log[(1-p)/p)]$, is $1$ at $p=0$, is $0$ at $p=1/2$, and has derivative 
$-2 (1 - 2 p) \log(1/p - 1)$, which is clearly negative.
A: By scaling if necessary, we may assume without loss of generality that $a^2p + b^2\bar p = 1$ Substituting $u = a^2p$, we can rewrite the 1-dimensional inequality as
\begin{align*}
f(u) := u\log \frac{u}{p} + \bar u \log \frac{\bar u}{\bar p} - c(p) (\sqrt{\bar u p} - \sqrt{u \bar p})^2 \le 0.
\end{align*}
We calculate the first two derivatives of $f(u)$. They are
\begin{align*}
f'(u) &= \log \frac{u}{\bar u} - \log \frac{p}{\bar p} - c(p)(1-2p) + c(p)(1-2u)\sqrt{\frac{p\bar p}{u\bar u}}
\end{align*}
and
\begin{align*}
f''(u) &= \frac{1}{u\bar u} - c(p)\frac{\sqrt{p\bar p}}{2(u\bar u)^{3/2}}.
\end{align*}
We first claim that $f''$ goes to $+\infty$ at 0 and 1, and precisely 2 zeros between 0 and 1 (they may be repeated). The first part is easy to check. To check the number of zeros, we need $2\sqrt{u\bar u} = c(p) \sqrt{p\bar p}$. As $u$ ranges from $[0,1]$, the left hand side ranges from $[0, 1]$ as well. It is not difficult to check that the right hand side is $\le 1$ (for example,  by differentiating it and checking that the derivative is 0 exactly at $p=1/2$, which is a maxima where the value is 1).
Having established this, we verify that $f'$ has the following zeros: $p, 1/2, \bar p$, and moreover, $f'$ is $+\infty$ at $0$ and $1$. Since $f''$ has exactly two zeros, $f'$ cannot have any more zeros. Recall that the points where $f'$ is 0 is precisely the list of maxima or minima of $f$. Moreover, the shape of $f''$ ensures that the local maxima are at $p$ and $\bar p$ whereas the local minimum is at $1/2$. Evaluating $f$ at the local maxima, we see that
\begin{align*}
f(p) = 0,
\end{align*}
and
\begin{align*}
f(\bar p) = \bar p \log \frac{\bar p}{p} + p \log \frac{p}{\bar p} - (\bar p - p) \log \frac{\bar p}{p} = 0.
\end{align*}
Thus, the maximum of $f$ is 0, and the result is proved.
