Exercise:
1. Prove that:
\[
\tag{1} 2\leq \frac{1}{\log_2 3} + \frac{1}{\log_3 2}\; .
\]
2. Can you provide any generalization of (1)?
gugo82 ha scritto:Ok, the first part is correct.
But look carefully: you don't really need Calculus to prove the inequality... The only tool you need is the elementary AMGM* inequality.Testo nascosto, fai click qui per vederloIn fact, inequality (1) rewrites:
\[
1\leq \frac{1}{2}\ \left( \frac{1}{\log_2 3} + \frac{1}{\log_3 2} \right)\; ,
\]
which is true because of the AMGM inequality:
\[
\sqrt{x\ y}=:\mathcal{G} (x,y)\leq \mathcal{A} (x,y):=\frac{1}{2}\ (x+y)
\]
and because:
\[
\mathcal{G} \left( \frac{1}{\log_2 3}, \frac{1}{\log_3 2} \right) = \sqrt{\frac{1}{\log_2 3}\ \frac{1}{\log_3 2}} = \sqrt{\frac{1}{\log_2 3}\ \log_2 3} =1\; .
\]
gugo82 ha scritto:On the other hand, the generalization you claim true for all positive real numbers does not work if $a=1$ nor if $b=1$... Thus you have to correct something in the last line of your previous post.
Visitano il forum: Nessuno e 1 ospite