Let $f:R\_+\to R\_+$ be smooth on $(0,\infty)$, increasing, $f(0)=0$ and 
$\lim\_{x\to\infty}=\infty$. Assume also that $f$ is subadditive:
$f(x+y)\le f(x)+f(y)$ for all $x,y\ge 0$. Must $f$ be concave? The converse is obvious.



subadditive implies concave
For not smooth surely not, take $f(x)=2x+|\sin x|$. I am nearly sure that for smooth answer is the same. For example, it looks like function $|\sin x|$ may be changed near points $\pi k$ so that it becomes smooth but still semiadditive.


well, more concrete construction is like follows (some details are however omited)


construct a function $f$ such that $f(x)=|\sin x|$ unless $|x-k\pi|<1/100$ for some positive integer $k$, $\sin 1/100\geq f(x) > \sin 1/100-1/1000000$ for $|x-k\pi| < 1/100$, $f$ is convex on $[k\pi-1.100,k\pi+1/100]$. When may $f(x+y) \le f(x)+f(y)$ fail? If $f(x+y)=|\sin(x+y)|$ then $f(x+y)\le |\sin x|+|\sin y|\le f(x)+f(y)$. If $f(x+y)\ne |\sin (x+y)|$, then $f(x+y)\le \sin(1/100)$, so if $f(x)+f(y) < f(x+y)$, then also $f(x) < \sin(1/100)$ and $f(y) < \sin(1/100)$. If both $x$ and $y$ are greater then $1/100$, then $f(x)+f(y) > 2(\sin(1/100)-1/1000000) > 1/100 > f(x+y)$.


Now without loss of generality $k\pi < y< x+y< k\pi+1/100$. Then $f(x+y)-f(y)=xf'(\theta)$ for some $\theta\in [y,x+y]$, by convexity $f'(\theta)\le \cos 1/100$. So, it suffices to prove that $x\cos(1/100)\le \sin x$, i.e. $\sin x/x\geq \cos 1/100$. Since $\sin t/t$ decreases on $[0,1/100]$, we have $\sin x/x\geq 100\sin 1/100\geq \cos 1/100$ as $\tan t > t$ for $t=1/100$


