Additive functional inequality The function $f:R_+\to R_+$ is continuously differentiable and
 increasing. Also, $f(0)=0$ and $f(\infty)=\infty$.   Continuity and
 differentiability of higher orders can be assumed if necessary.   The
 proposition on hand is the following:


If for all integers $t>0$ and
    for all  $x> 0$, $f((t+1)x)<f(tx)+f(x)$, then for all
    integers $m,n>0$, $f(mx)+f(nx)\geq f(mx+nx)$. 


Does there exist a proof or a counter example?
 A: The following is a counterexample:
We can write
$$f(x) = \int_0^xf'(\xi)d\xi$$
If we write $g(x)$ for $f'(x)$ we can restate the problem in terms of $g$: $g$ is non-negative everywhere, it cannot go to $0$ too fast (so that its integral tends to $\infty$), and
$$\int_{kx}^{(k+1)x}g(\xi)d\xi < \int_{0}^{x}g(\xi)d\xi$$
The question is if then also for all integers $m,n > 0$ we must have that
$$\int_{mx}^{(m+n)x}g(\xi)d\xi < \int_{0}^{nx}g(\xi)d\xi$$
To construct the counterexample, first consider the function $G$ that, for some small number $\delta>0$, goes linearly from $10+\delta$ to $10$ between $0$ and $1$, is $0$ up to $2$, then constantly $6$ up to $5/2$, $0$ up to $3$, $6$ up to $5$ and from there some small constant value $\varepsilon$ to ensure its integral will tend to infinity. 

To ease notation, write
$$I[a,b] = \int_a^bG(\xi)d\xi$$
(so that $f(x) = I[0,x]$).  
First we have to show that for all $x>0$ and all $k$
$$I[0,x] > I[kx, (k+1)x]$$
For $x \le 1$ that is obviously true, and actually up to 5/3 the average value of $G(x)$ on $[0,x]$ is more than 6, which it is on no interval starting after 1, so that for $x\le 5/3$ it is OK. 
Next consider the case that $x = 2 - t$, for $0 < t < 1/3$. Then $I[0,x] = 10 + \delta/2$, $I[x,2x] < 9$, $I[2x,3x] = 6(1 + 2t) \le 10$ and for other $I[kx, (k+1)x]$ we are good. 
You can treat the case $x = 2 + t$, for $0 < t < 1/2$ similarly, and for larger $x$ it is easy to see that it will work if you make $\varepsilon$ small enough (or you make $G$ zero for a while before making it a nonzero constant).
However, note that we have
$$I[0,2] < I[3,5]$$
so that
$$f(x) = \int_0^xG(\xi)d\xi$$
almost satisfies the conditions, but is a counterexample for $x = 1$, $m = 3$, $n = 2$. 
I say almost, because the integral of $G$ is not continuously differentiable. We do however have enough freedom in the inequalities to smoothen $G$ to a function $g$ without violating them, and if we would want to make $f$ strictly increasing we could add a constant to $g$.
EDIT
To make the smoothing explicit you could explicitly replace the jumps by sufficiently steep slopes, but maybe the following is more elegant, and gives a $\mathcal C^\infty$-function:
Consider a sequence $(\zeta_n)$ of positive $\mathcal C^\infty$-functions with $\int_{-\infty}^\infty\zeta_n(x)dx = 1$ that converge to a Dirac mass $\delta_0$ centered at $0$ in the sense of distributions. Convolving with $\delta_0$ is the same as not doing anything, so the sequence of convolutions $g_n := G\ast\zeta_n$, which are $\mathcal C^\infty$-functions, converges to $G$. As an example you could take $\zeta_n(x) = {1\over n\sqrt{2\pi}}e^{-{x^2\over 2n^2}}$.
Essentially this means that all integrals of the $g_n$ converge to the integrals of $G$, and the required inequalities will hold for $g_n$ already, for sufficiently large $n$. 
First note that $g_n$ is decreasing on $[0,1]$, so the required inequality still holds for $x\le 1$. For $x \le 5/3$ we now have to fix $\delta$ first, because $n$ will depend on it (for very small $\delta$ you may have to take a very large $n$), but we can do the same thing as before.
Likewise for the other inequalities, which are strict. Since there are essentially only finitely many of them, of integrals on subsets of a bounded set, it is clear that the required $n$ can be chosen uniformly. 
Finally, let $g := g_n$ for this $n$. Then 
$$f(x) = \int_0^xg(\xi)d\xi$$
is a $\mathcal C^\infty$-function satisfying all properties but not the conclusion, i.e. it is a counterexample.
