0

How to prove:

$$\log_{a}{\frac{a^n+b^n}{a^m+b^m}}+\log_{b}{\frac{a^n+b^n}{a^m+b^m}} \geq 2(n-m),$$ where $n>m$, $a,b \in (1, \infty)$. I tried some methods such as

$$ a^n +b^n \leq (a+b)^n$$ but with no result, at least not right.

or $$ \log_{a}{(a^n+b^n)}-\log_{a}{(a^m+b^m)} \geq n-m $$ $$ \log_{b}{(a^n+b^n)}-\log_{b}{(a^m+b^m)} \geq n-m $$ And then I used $$\log(x+y)=\log(x)+\log(1+\frac{y}{x})$$ Maybe it's a good start, but I don't know how to continue.

ACB
  • 3,713
Mark Ben
  • 181
  • 1
    Have you tried using the log rule to get rid of the fractions and examining what you get from that? – Mark Bennet Oct 16 '21 at 11:59
  • I tried, but it gets very complicated and I don't have any idea about how to simply and to arrive at 2$(n-m)$. – Mark Ben Oct 16 '21 at 12:36

1 Answers1

1

Using quotient rule of logarithm we have, $$\log_{a}(a^{n}+b^{n})+\log_{b}(a^{n}+b^{n})-(log_{a})(a^{m}+b^{m})+\log_{b}(a^{m}+b^{m})$$ For all $x,y,z\in\mathbb{R^{+}} x^{k}\le(x^{k}+y^{k})\le(x+y)^{k}$ Using that it can be showed that $$n+n-m(\log_{a}(a+b)-\log_{b}(a+b)\le\log_{a}(a^{n}+b^{n})+\log_{b}(a^{n}+b^{n})-(log_{a}(a^{m}+b^{m})+\log_{b}(a^{m}+b^{m})$$ $$n+n-m(2)\le n+n-m(\log_{a}(a+b)-\log_{b}(a+b)$$

RAHUL
  • 1,511