Prove that $\log_n(n+1)\geq\log_{n+1}(n+2)$ for $n>1$.
So far I only know that $\log_n(n+1)>\frac{\log_n(n+2)}{\log_n(n+1)}$ Since $n>1$, LHS must be greater than RHS.
Is there any other ways to prove this inequality. A proof for this inequality can be found here: https://brilliant.org/wiki/logarithmic-inequalities/
The proof is incorrect since the author confuses $(\log(n+1))^2$ for $\log(n+1)^2=\log(n^2+2n+1)$. They are different.
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HINT:
Consider the function $$f(x)=\log_x(x+1)=\frac{\log(x+1)}{\log x}.$$ Then $$f'(x)=\frac{\frac{\log x}{x+1}-\frac{\log(x+1)}x}{\log^2x}=\frac{x\log x-(x+1)\log(x+1)}{x(x+1)\log^2x}$$ and since $x$ is positive and $\log$ is monotonically increasing, what is the sign of $f'(x)$?
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By AM-GM $$\ln_n(n+1)-\ln_{n+1}(n+2)=\frac{1}{\ln_{n+1}n}-\ln_{n+1}(n+2)=$$ $$=\frac{1-\ln_{n+1}n\ln_{n+1}(n+2)}{\ln_{n+1}n}\geq\frac{1-\left(\frac{\ln_{n+1}n+\ln_{n+1}(n+2)}{2}\right)^2}{\ln_{n+1}n}=$$ $$=\frac{1-\frac{1}{4}\ln^2_{n+1}(n^2+2n)}{\ln_{n+1}n}>\frac{1-\frac{1}{4}\ln^2_{n+1}(n^2+2n+1)}{\ln_{n+1}n}=0.$$
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By the change of base formula \begin{align} \log_b x = \frac{\log_a x}{\log_a b} \end{align} we see that \begin{align} \log_{n+1}(n+2) = \frac{\log_n (n+2)}{\log_n(n+1)}. \end{align}
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The proof is incorrect since the author confuses $(\log(n+1))^2$ for $\log(n+1)^2=\log(n^2+2n+1)$. They are different.
No, the proof is correct. It does not directly substitute one for another, but it uses estimates. The main inequality to be proved is: $(\log(n+1))^2>\log n\log(n+2)$.
It uses the AM-GM (arithmetic mean-geometric mean) to show: $$(\log(n+1))^2=\left(\frac{\log(n+1)^2}{2}\right)^2>\left(\frac{\log (n(n+2))}{2}\right)^2=\\ \left(\frac{\log n +\log (n+2))}{2}\right)^2\stackrel{AM-GM}{>}\log n\log(n+2).$$
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You are correct $( \ln(n+1))^2 \neq \ln((n+1)^2)$. But the latter expression is crucial in showing the ineqaulity. \begin{eqnarray*} \ln(n+2)-\ln(n+1) \leq \ln(n+1)-\ln(n) \end{eqnarray*} now multiply by $\ln(n) \leq \ln(n+1)$ to obtain the required inequality.