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?



Exercise related to log-Sobolev inequalities
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.


