# 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!

• Please check $a = 1, b = 2, p = 5/6$. Commented Dec 31, 2021 at 2:52

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...

• I guess: If $a^2 < b^2$, then $f(p) \le 0$ for all $0 < p < 1/2$; If $a^2 > b^2$, then $f(p) \le 0$ for all $1/2 < p < 1$. Commented Dec 31, 2021 at 6:29
• Are there any intuitions behind your guess? Commented Dec 31, 2021 at 7:32
• Let $c = b^2/a^2$. Did some numerical experiment. Commented Dec 31, 2021 at 9:51

Hints.

Consider this:

$$\lim_{p\to 0} f(p) = -\infty$$ $$\lim_{p\to 1} f(p) = 0$$

$$\lim_{p\to 0^+} f'(p) = +\infty$$ $$\lim_{p\to 1} f'(p) = +\infty$$
$$\lim_{p\to 0} f''(p) = -\infty$$ $$\lim_{p\to 1^-} f''(p) = +\infty$$