TITLE: Convexity of $x\left(1+\frac1x\right)^x,\ x\ge 0$
QUESTION [16 upvotes]: This may turn out to be really simple but I do not see a quick way to the proof. How would one show $\displaystyle x\Big(1+\frac1x\Big)^x,\ x\ge 0$ is convex?
I derived the second derivative. It has a negative term. I suppose I could combine certain terms to make the negativeness disappear. But I hope there is a really clever way to see it right away.

REPLY [4 votes]: I just want to add to the elegant proof of @wangtwo, and remedy the only blemish of an ugly proof of the convexity of $\frac{\ln(1+x)}x$ in a piece of otherwise perfect work. Here is an elegant replacement.
$$\frac{\ln(1+x)}x=\int_0^1 \frac1{1+xt}dt$$
is convex since the integrand is convex with respect to $x$.

Here is a closely related problem and my own solution as mentioned by @wangtwo in his comment below.