Convexity of $x\left(1+\frac1x\right)^x,\ x\ge 0$ 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.

Let $f(x)=x\left(1+\frac{1}{x}\right)^x$, where $x>0$.
Hence,
$$f''(x)=\frac{\left(1+\frac{1}{x}\right)^x\left(x(x+1)^2\ln^2\left(1+\frac{1}{x}\right)+2(x+1)\ln\left(1+\frac{1}{x}\right)-x-3\right)}{(x+1)^2}.$$
Thus, we need to prove that 
$$\ln\left(1+\frac{1}{x}\right)>\frac{\sqrt{x^2+3x+1}-1}{x(x+1)}.$$
Let $g(x)=\ln\left(1+\frac{1}{x}\right)-\frac{\sqrt{x^2+3x+1}-1}{x(x+1)}.$
Hence, $$g'(x)=\frac{2x^3+9x^2+7x+2-2\sqrt{(x^2+3x+1)^3}}{2x^2(x+1)^2\sqrt{x^2+3x+1}}=$$
$$=-\frac{11x^3+46x^2+35x+8}{2x(x+1)^2\sqrt{x^2+3x+1}\left(2x^3+9x^2+7x+2+2\sqrt{(x^2+3x+1)^3}\right)}<0,$$
which says that $g(x)>\lim\limits_{x\rightarrow+\infty}g(x)=0$ and we are done!
