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Let $f(x) = (x-1)\ln x$, and given $0 < a < b$.

If $f(a) = f(b)$, how to prove that $\frac{1}{\ln a}+\frac{1}{\ln b} < \frac{1}{2}$?

ShBh
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xunitc
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  • This one has kept me going for quite a while. Playing around in DESMOS, is it possible that the $1/2$ should have an inclusion sign? I think this is a great question, wondering why it has had so few "viewers" – imranfat May 23 '18 at 13:57
  • @imranfat Thank you for your attention. it just $<$, not $\le$. – xunitc May 24 '18 at 09:27
  • Well, I am keen to see an answer, and since you tagged it without Calculus, I am even more interested in an answer! With programming I can see it work, but that ain't no proof – imranfat May 24 '18 at 18:05
  • Wait, if $b>a \Rightarrow \ln b > \ln a\ &\ (b-1)>(a-1) \Rightarrow (b-1) \ln b > (a-1) \ln a \Rightarrow f(b)>f(a)$? – Suhrid Saha Oct 24 '18 at 18:26
  • @SuhridSaha the second $\Rightarrow$ is not right. because $a-1$ and $\ln a$ can both $< 0$. You can plot the pic of $f(x)$, then you will see the $a,b$ there. – xunitc Oct 25 '18 at 09:26
  • @xunitc Oh yeah, i really make such stupid mistakes :P – Suhrid Saha Oct 25 '18 at 14:49

1 Answers1

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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:

  1. $b \ge \mathrm{e}^2$: Then $\frac{1}{2}(b-1)\ln b - b + 2 \ge 1$. Clearly, (1) is true.

  2. $1 < b < \mathrm{e}^2$: 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 $(0, 2)$.

We are done.

River Li
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