A Logarithmic Inequality

[gugo82]

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)?

[Sk_Anonymous]

"gugo82":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.

In 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:
\[
\mathcal{G} \left( \frac{1}{\log_2 3}, \frac{1}{\log_3 2} \right) = \sqrt{\frac{1}{\log_2 3}\ \frac{1}{\log_3 2}} = \sqrt{\frac{1}{\log_2 3}\ \log_2 3} =1\; .
\]

Well, I just wrote the very first proof it came into my mind as soon as I saw the inequality.
This simple and elementary one is much better

"gugo82":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.

Ups, you're right. I suppose it is enough to assume $a$ and $b$ positive real numbers different from $1$.

[gugo82]

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.

In 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:
\[
\mathcal{G} \left( \frac{1}{\log_2 3}, \frac{1}{\log_3 2} \right) = \sqrt{\frac{1}{\log_2 3}\ \frac{1}{\log_3 2}} = \sqrt{\frac{1}{\log_2 3}\ \log_2 3} =1\; .
\]

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.

__________
* AM stands for arithmetic mean, while GM for geometric mean.

[Sk_Anonymous]

First of all, sorry for my bad english.

$1/{\log_ab}+1/{\log_ba}=1/{\log_ab}+\log_ab$

Let $f(x)$ be the function $f(x)=x+x^{-1}$. We have $\lim_{x->0^+}f(x)=+oo$ and $\lim_{x->0-}f(x)=-oo$.
Its first derivative is $f\ '(x)=1-x^{-2}$, which equals $0$ for $x=1$ or $x=-1$.
Its second derivative is $f\ ''(x)=2x^{-3}$ so $f(x)$ has a maximum at $x=-1$ (and $f(-1)=-2$) and a minimum at $x=1$ (and $f(1)=2$).
So if $x>0$ we have $f(x)>=2$ and if $x<0$ we have $f(x)<=-2$. In any case $|f(x)|>=2$

$1/{\log_2 3}+1/{\log_3 2}=f(\log_3 2)>=2$

If $a$ and $b$ are positive real numbers then $|1/{\log_ab}+1/{\log_ba}|=|f(\log_ab)|>=2$

[gugo82]

Thanks Camillo.
It's always a pleasure for me to contribute to this section of our wonderful forum.

[Camillo]

Gugo welcome back