<s>[INST] $(\frac{x}{x+1})^x$ decreasing 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?
 [/INST] 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
$$1+(x+1)\log\left(\frac{x}{x+1}\right) < 1 + (x+1)\cdot\frac{-1}{x+1} = 0.$$
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