In my math textbook, I came across this question which I do not know how to answer. For a variable n, where n>1, how would you prove that $\log_n (n+1) > \log_{n+1} (n+2)$?
Thank you.
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What I would do is take the limit of $\log_n(n+1)$ as n tends to $\infty$ and try making something out of that. If the limit converges to $0$, then the inequality is correct, if it diverges to $\infty$, then the inequality is wrong. – Badr B Nov 28 '17 at 02:53
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I wonder if it is true that $\log_x(x+1)$ is a strictly decreasing function on $(1,\infty)$? – MPW Nov 28 '17 at 02:53
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Consider $f(t) = \log_t(t+1)={\log(t+1)\over\log(t)}$, so $$f'(t) = \frac{\frac{1}{t+1} \log (t) - \frac{1}{t} \log(1+t)}{\log(t)^2}$$ But $x \log x$ is an increasing function on $x>1$, so $t \log t < (t+1) \log(t+1)$, hence $f'(t) <0$ on $(1, \infty)$ so $f$ is decreasing, as required.
B. Mehta
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it can be shown to be true also without calculus by AM-GM inequality
$$\frac{a+b}{2} \geq \sqrt{ab}$$
firstly observe that
$$\log_n (n+1) = \frac {\log (n+1)}{\log(n)}$$ $$\log_{n+1} (n+2) = \frac {\log (n+2)}{\log(n+1)}$$
then
$$\log_n (n+1) > \log_{n+1} (n+2) \iff \log^2 (n+1) > \log(n)\log(n+2)$$
now for AM-GM inequality
$$\log(n)\log(n+2)\leq \left( \frac{log(n)+log(n+2)}{2} \right)^2 = \left( \frac{log(n^2+2n)}{2} \right)^2 < \left( \frac{log(n^2+2n+1)}{2} \right)^2 = \log^2 (n+1)$$
user
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