$(\frac{x}{x+1})^x$ decreasing

Question

I want to show that $f(x) = \left(\dfrac{x}{x+1}\right)^x$ is decreasing for $x > 0$; this is clear from plotting its graph. Taking its derivative,

$$f'(x) = \frac{x^x}{(x+1)^{x+1}}\left(1+(x+1)\log\left(\frac{x}{x+1}\right)\right).$$

So we need to show that $1+(x+1)\log\left(\frac{x}{x+1}\right) < 0$, which seems to require taking another derivative. Is there an easier way to show that $f(x)$ is decreasing?

just use $\ln(1+x) < x$ – YOu will not know Jun 29 '21 at 15:09 — YOu will not know, Jun 29 '21 at 15:09

score 3 · Answer 1 · answered Jun 29 '21 at 15:13

3

Use the fact that $$ \left(\frac{x}{x+1}\right)^x=\frac{1}{\left(1+\frac{1}{x}\right)^x} \, . $$ Does the denominator look familiar to you?

answered Jun 29 '21 at 15:13

Joe

19,636

Ah, yes, its limit is $e$... but now, why is the denominator increasing (without taking the derivative again)? – tarthoe Jun 29 '21 at 15:16
1

@tarthoe: Does this question help? – Joe Jun 29 '21 at 15:22
At the link you gave, I added an answer https://math.stackexchange.com/questions/83035/how-to-prove-11-xx-is-increasing-when-x0/4186249#4186249. – zhw. Jun 29 '21 at 16:58
@tarthoe: See zhw's answer—I like it very much. – Joe Jun 29 '21 at 19:26
Thank you to both! – tarthoe Jun 30 '21 at 01:25
1

@tarthoe: I'm glad I could help. Let me know if you have any questions. – Joe Jun 30 '21 at 11:59

score 2 · Accepted Answer · answered Jun 29 '21 at 15:28

By taking a derivative, you can easily show that

$$\log(1+y) \le y$$

for all $y > -1$, with equality iff $y=0$ (this is a standard inequality). In particular, since $-1<-\frac{1}{x+1}<0$,

$$\log\left(\frac{x}{x+1}\right) = \log\left(1 - \frac{1}{x+1}\right) < - \frac{1}{x+1}.$$

Therefore

$$1+(x+1)\log\left(\frac{x}{x+1}\right) < 1 + (x+1)\cdot\frac{-1}{x+1} = 0.$$

$(\frac{x}{x+1})^x$ decreasing

2 Answers2