Jensen-like averaging inequality on integers Let $\mathbb{Z}^*=\mathbb{Z}^+\cup\{0\}$. Let $f:\mathbb{Z}^*\rightarrow\mathbb{R}$ be a nondecreasing function such that $f(a+b)\leq f(a)+f(b)$ for all $a,b\in\mathbb{Z}^*$. 
Is it true that for all $k,n\in\mathbb{Z}^+$, we have $$f(k)\ge\frac{f(a_1)+\dots+f(a_n)}{2n}$$ for all $a_1,\dots,a_n\in\mathbb{Z}^*$ with $a_1+\dots+a_n=kn$?
For $n=2$, this is true, since $f(k)\geq\frac{f(2k)}{2}\geq\frac{f(a_1)+f(a_2)}{4}$, where we use both conditions on $f$. For $n=1$ it is also true trivially.

The statement is true, even with strict inequality.
By induction we have $f(nk)\leq nf(k)$. Suppose that we round each $a_i$ up to the next multiple of $k$, say $b_i$, so $f(a_i)\leq f(b_i)$ where $b_i-a_i<k$. Since $\sum_{i=1}^na_i=kn$, we have $\sum_{i=1}^nb_i<2kn$. Thus,
$$\sum_{i=1}^nf(a_i)\leq\sum_{i=1}^nf(b_i)\leq f(k)\sum_{i=1}^n\frac{b_i}{k}<2nf(k).$$ 
