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This question already has an answer here:

Please tell me how to solve this one. Given that $m, n$ are positive real numbers and $m>n>0$. Prove that $(1+\frac1m)^m > (1+\frac1n)^n$.

Now I am willing to prove this through elementary inequality techniques( like AM-GM inequality etc). I know that the same can be proved using calculus and the fact $\{(1+\frac1x)^x: x >0\}$ is an increasing function.

But without using the calculus stuff, can we prove it in simple inequality way ? Thanks in advance

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marked as duplicate by Martin R, José Carlos Santos, Lord Shark the Unknown, JonMark Perry, Claude Leibovici Jul 24 '17 at 6:15

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  • $\begingroup$ Why are you talking about negative integers when $m,n>0$? What's your easy proof when $m$ and $n$ are positive integers? $\endgroup$ – Ted Shifrin Jul 24 '17 at 2:52
  • $\begingroup$ My bad. I didn't notice it was given $m,n$ are positive. For positive integers $m,n$ I think we can do the following: First we show the sequence $\{(1+\frac1n)^n\}$ is increasing sequence. Then since, $m>n$ we\rq ll be done. For first part, we consider $n+1$ positive numbes $1+frac1n, 1+\frac1n, \cdots, 1+\frac1n (n times)$ and 1. Applying AM>GM we shall get $(1+\frac{1}{n+1})^{n+1}>(1+\frac{1}{n})^n$\ $\endgroup$ – Anjan3 Jul 24 '17 at 2:59
  • $\begingroup$ Also: math.stackexchange.com/q/1011414/42969, math.stackexchange.com/q/167843/42969. $\endgroup$ – Martin R Jul 24 '17 at 4:13
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Hint It is enough to show that

$$\left(1+ \frac{1}{n+1}\right)^{n+1} > \left( 1+\frac{1}{n} \right)^n$$

Hint 2 $$\frac{\left(1+ \frac{1}{n+1}\right)^{n+1} }{ \left( 1+\frac{1}{n} \right)^{n+1}}=\left(1-\frac{1}{(n+1)^2} \right)^{n+1}> \left(1-\frac{n+1}{(n+1)^2} \right) $$

with the inequality following from Bernoulli.

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