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 best I can do is with $n^2$ in the denominator.
By induction, $f(kn) \leq \underbrace{f(k)+\cdots+f(k)}_{n}$, so $f(k) \geq \frac{f(kn)}{n}$. as $f$ is nondecreasing, $f(kn) \geq f(a_i)$ for $1 \leq i \leq n$. Hence $nf(kn) \geq f(a_1)+\cdots +f(a_n)$, and thus $f(kn) \geq \frac{f(a_1)+\cdots+f(a_n)}{n}$. Combining this with the previous inequality, $f(k) \geq \frac{f(a_1)+\cdots +f(a_n)}{n^2}$. 
