Show $2 + \log\left(\frac{a^2}{pa^2 + (1-p)b^2}\right) - \frac{2pa^2}{pa^2 + (1-p)b^2} - (1-p)\frac{1}{1-2p}\log\frac{1-p}{p} \leq 0$ for $p\in [0,1]$

My question is related to the proof of Theorem 5.2 in this monograph (see also Exercise 5.4). Specifically, following the argument of the Theorem 5.1, I am guessing that $$f(p): = 2 + \log\left(\frac{a^2}{pa^2 + (1-p)b^2}\right) - \frac{2pa^2}{pa^2 + (1-p)b^2} - (1-p)\frac{1}{1-2p}\log\frac{1-p}{p} \leq 0 ~~\textrm{for all}~~ p\in [0,1].$$ The inequality $f(1/2) \leq 0$ is just a consequence of the limiting behavior $\lim_{p \to 1/2} \frac{1}{1-2p}\log\frac{1-p}{p} = 1$ and the usual inequality $\log(x) \leq x - 1$ for all $x >0$. But I have no clue as to show that $f(p) \leq 0$ for $p\neq 1/2$. I appreciate any kind help on this problem!

Solution 1:

Thanks to the suggestion given by @River Li, consider the case where $a = 1$, $b = 2$, and $ p = 5/6$. Then $$f(p) = 2 + \log(3/2) - \frac{10}{9} - \frac{1}{4}\log(6) \approx 0.8464 > 0.$$ Therefore, the conjecture is wrong! The message is that the proof strategy used for the proof of Theorem 5.1 does not carry over to the proof of Theorem 5.2, which is a pity...