let $f(x) = (x-1)\ln x$, and given $0 < a < b$. If $f(a) = f(b)$, prove that $\frac{1}{\ln a}+\frac{1}{\ln b} < \frac{1}{2}$

Proof.

Let $f(x) = (x-1)\ln x$. Clearly, $f(1) = 0$, $\lim_{x\to 0^{+}} f(x) = \infty$, and $\lim_{x\to \infty} = \infty$. We have $f'(x) = \ln x + 1 - \frac{1}{x}$. Clearly, $f(x)$ is strictly decreasing on $(0, 1)$, and strictly increasing on $(1, \infty)$. Thus, $0 < a < 1$ and $1 < b$.

We have $$ \frac{1}{\ln a} + \frac{1}{\ln b} = \frac{a-1}{(a-1)\ln a} + \frac{b-1}{(b-1)\ln b} = \frac{a+b-2}{(b-1)\ln b}. $$ It suffices to prove that $$a < \frac{1}{2}(b-1)\ln b - b + 2. \tag{1}$$ We split into two cases:

$b \ge \mathrm{e}^2$: Then $\frac{1}{2}(b-1)\ln b - b + 2 \ge 1$. Clearly, (1) is true.
$1 < b < \mathrm{e}^2$: It is not difficult to prove that $0 < \frac{1}{2}(b-1)\ln b - b + 2 < 1$. Since $f(x)$ is strictly decreasing on $(0, 1)$, it suffices to prove that $$f(a) > f(\tfrac{1}{2}(b-1)\ln b - b + 2)$$ or $$f(b) > f(\tfrac{1}{2}(b-1)\ln b - b + 2)$$ or $$(b-1)\ln b - \left(\tfrac{1}{2}(b-1)\ln b - b + 1\right)\ln \left( \tfrac{1}{2}(b-1)\ln b - b + 2 \right) > 0$$ or $$\frac{1}{2}(b-1)\left[ 2\ln b + (2 - \ln b)\ln \left(1 - \tfrac{1}{2}(b-1)(2-\ln b)\right)\right] > 0.$$ It suffices to prove that $$2\ln b + (2 - \ln b)\ln \left(1 - \tfrac{1}{2}(b-1)(2-\ln b)\right) > 0.$$ Let $b = \mathrm{e}^y$. Then $0 < y < 2$. It suffices to prove that $$2y + (2-y)\ln \left(1 - \tfrac{1}{2}(\mathrm{e}^y-1)(2-y)\right) > 0$$ or $$\frac{2y}{2-y} + \ln \left(1 - \tfrac{1}{2}(\mathrm{e}^y-1)(2-y)\right) > 0.$$ Denote the LHS by $g(y)$. We have $$g'(y) = \frac{(y^3-5y^2+12y-12)\mathrm{e}^y - y^2 + 12}{(2-y)^2(2 - (\mathrm{e}^y-1)(2-y))}.$$ It is not difficult to prove that $(y^3-5y^2+12y-12)\mathrm{e}^y - y^2 + 12 > 0$ for all $y$ in $(0, 2)$. Thus, $g'(y) > 0$ for all $y$ in $(0, 2)$. Also, $g(0) = 0$. Thus, $g(y) > 0$ for all $y$ in $(0, 2)$.