Find infimum or minimum of $\frac{\log_{a} b}{a-b+1}+\frac{\log_{b} c}{b-c+2}+\frac{\log_{c} a}{c-a+3}$

Remarks: The proof of (2) is quite complicated. We hope to see a nice proof.

Fact 1: Let $a, b, c \in (0, 1)$. Then $$\frac{\log_{a} b}{a-b+1}+\frac{\log_{b} c}{b-c+2}+\frac{\log_{c} a}{c-a+3} > \frac12\sqrt[3]{36}.$$ (The proof is given at the end.)

On the other hand, letting ($z > 0$) $$a = \mathrm{exp}\left(-\frac{\sqrt[3]{36}}{2z}\right), \quad b = \mathrm{exp}\left(-\frac{\sqrt[3]{6}}{2z}\right), \quad c = \mathrm{exp}\left(-\frac{1}{z}\right),$$ we have $a, b, c\in (0, 1)$, and $$\lim_{z\to \infty} \left(\frac{\log_{a} b}{a-b+1}+\frac{\log_{b} c}{b-c+2}+\frac{\log_{c} a}{c-a+3}\right) = \frac12\sqrt[3]{36}. \tag{1}$$

Using Fact 1 and (1), we have $$\inf_{a, b, c\in (0, 1)} \left(\frac{\log_{a} b}{a-b+1}+\frac{\log_{b} c}{b-c+2}+\frac{\log_{c} a}{c-a+3}\right) = \frac12\sqrt[3]{36}.$$ (Note: The infimum is not attained.)

Proof of Fact 1:

Using Cauchy-Bunyakovsky-Schwarz inequality, we have \begin{align*} &\left(\frac{\log_{a} b}{a-b+1}+\frac{\log_{b} c}{b-c+2}+\frac{\log_{c} a}{c-a+3}\right) \left((a - b + 1) + \frac{b - c + 2}{2} + \frac{c - a + 3}{3}\right)\\ \ge\,\,& \left(\sqrt{\log_{a} b} + \sqrt{\frac{\log_{b} c}{2}} + \sqrt{\frac{\log_{c} a}{3}}\right)^2. \end{align*} Also, we have \begin{align*} &\left(\sqrt{\log_{a} b} + \sqrt{\frac{\log_{b} c}{2}} + \sqrt{\frac{\log_{c} a}{3}}\right)^2\\ =\,\,& \log_{a} b + \frac{\log_{b} c}{2} + \frac{\log_{c} a}{3} + 2 \sqrt{\log_{a} b} \sqrt{\frac{\log_{b} c}{2}} + 2 \sqrt{\frac{\log_{b} c}{2}} \sqrt{\frac{\log_{c} a}{3}} + 2\sqrt{\frac{\log_{c} a}{3}} \sqrt{\log_{a} b}\\ \ge\,\,& \log_{a} b + \frac{\log_{b} c}{2} + \frac{\log_{c} a}{3} + 3\sqrt[3]{2 \sqrt{\log_{a} b} \sqrt{\frac{\log_{b} c}{2}} \cdot 2 \sqrt{\frac{\log_{b} c}{2}} \sqrt{\frac{\log_{c} a}{3}} \cdot 2\sqrt{\frac{\log_{c} a}{3}} \sqrt{\log_{a} b}}\\ =\,\,& \log_{a} b + \frac{\log_{b} c}{2} + \frac{\log_{c} a}{3} + \sqrt[3]{36}. \end{align*} Thus, we have $$\frac{\log_{a} b}{a-b+1}+\frac{\log_{b} c}{b-c+2}+\frac{\log_{c} a}{c-a+3} \ge \frac{\log_{a} b + \frac{\log_{b} c}{2} + \frac{\log_{c} a}{3} + \sqrt[3]{36}}{\frac23 a - \frac12 b - \frac16 c + 3}. $$

It suffices to prove that $$\log_{a} b + \frac{\log_{b} c}{2} + \frac{\log_{c} a}{3} > \frac12 \sqrt[3]{36}\left(\frac23 a - \frac12 b - \frac16 c\right) + \frac12 \sqrt[3]{36}. \tag{2}$$

We split into two cases:

Case 1: If $4a < 3b + c$, we have $\frac23 a - \frac12 b - \frac16 c < 0$. Also, we have $$\log_{a} b + \frac{\log_{b} c}{2} + \frac{\log_{c} a}{3} \ge 3\sqrt[3]{\log_{a} b \cdot \frac{\log_{b} c}{2} \cdot \frac{\log_{c} a}{3}} = \frac12 \sqrt[3]{36}.$$ Thus, (2) is true.

Case 2: If $4a \ge 3b + c$, using $\frac12 \sqrt[3]{36} < 5/3$ and $\frac23 a - \frac12 b - \frac16 c \ge 0$, it suffices to prove that $$\log_{a} b + \frac{\log_{b} c}{2} + \frac{\log_{c} a}{3} > \frac{5}{3}\left(\frac23 a - \frac12 b - \frac16 c\right) + \frac{5}{3}.$$ Note that $$ \frac13 \log_a b + \frac13 \log_b c + \frac13 \log_c a \ge 3\sqrt[3]{\frac13 \log_a b \cdot \frac13 \log_b c \cdot \frac13 \log_c a} = 1. $$ It suffices to prove that $$\frac23 \log_{a} b + \frac16 \log_{b} c + 1 > \frac53\left(\frac23 a - \frac12 b - \frac16 c\right) + \frac53. \tag{3}$$

Fact 2: If $1 > x \ge y > 0$, then $\log_x y \ge \frac53(x - y) + 1$.
(Proof: Let $f(y) = \log_x y - \frac53(x - y) - 1$. We have $f''(y) = \frac{1}{y^2\ln(1/x)} > 0$ for all $y\in (0, 1)$. Using the convexity of $f$, we have $f(y) \ge f(x) + f'(x)(y - x) = (\frac{1}{x\ln x} + \frac53)(y - x) \ge 0$ where we have used $-\mathrm{e}^{-1}\le x\ln x < 0$ for all $x\in (0, 1)$.)

We split into three cases:

Case 2.1: If $a \ge b$ and $c \ge b$, (3) is written as $$\frac23\left(\log_a b - \frac53(a - b) - 1\right) + \frac16 \log_b c + \frac{5}{18}(c - b) > 0. \tag{4} $$

Using Fact 2, (4) is true.

Case 2.2: If $a \ge b \ge c$, (3) is written as $$\frac23\left(\log_a b - \frac53(a - b) - 1\right) + \frac16\left(\log_b c - \frac53(b - c) - 1\right) + \frac16 > 0. \tag{5} $$

Using Fact 2, (5) is true.

Case 2.3: If $b > a \ge c$, it suffices to prove that $$\frac23 \log_{a} b + \frac16 \log_{b} c + 1 \ge \frac53\left(\frac23 a - \frac12 a - \frac16 c\right) + \frac53. $$ Using Fact 2, we have $$\frac16 (\log_a c - 1) \ge \frac53\left(\frac23 a - \frac12 a - \frac16 c\right).$$ It suffices to prove that $$\frac23 \log_{a} b + \frac16 \log_{b} c + 1 \ge \frac16 (\log_a c - 1) + \frac53 $$ or $$\frac23 \log_{a} b + \frac16 \log_{b} c - \frac16 \log_a c - \frac12 \ge 0$$ or $$\frac{4\ln \frac{1}{b} - \ln \frac{1}{c}}{6\ln \frac{1}{a}} + \frac{\ln \frac{1}{c}}{6\ln\frac{1}{b}} - \frac12 \ge 0.$$

We split into two cases:

Case 2.3.1: If $4\ln \frac{1}{b} - \ln \frac{1}{c} < 0$, it suffices to prove that $$\frac{4\ln \frac{1}{b} - \ln \frac{1}{c}}{6\ln \frac{1}{b}} + \frac{\ln \frac{1}{c}}{6\ln\frac{1}{b}} - \frac12 \ge 0$$ which is clearly true.

Case 2.3.2: If $4\ln \frac{1}{b} - \ln \frac{1}{c} \ge 0$, it suffices to prove that $$\frac{4\ln \frac{1}{b} - \ln \frac{1}{c}}{6\ln \frac{1}{c}} + \frac{\ln \frac{1}{c}}{6\ln\frac{1}{b}} - \frac12 \ge 0$$ or $$\frac23 \log_c b + \frac16 \log_b c \ge \frac23$$ or $$\frac{(2\log_c b - 1)^2}{6\log_c b} \ge 0$$ which is clearly true.

We are done.