I'm looking for a 1-homogeneous function $\pi \colon \mathbb{R}^n_{\geq 0} \to \mathbb{R}$ satisfying the following properties:

1) $\pi$ is not concave. This is equivalent to the fact that there exist $x, y \in\mathbb{R}^n_{\geq 0}$ such that $\pi(x+y) < \pi(x) + \pi(y)$.

2) For all $v = (v_1, \cdots , v_n) \in\mathbb{R}^n_{\geq 0}$ the quantity $\frac{i}{n}! \frac{\pi(v)^n}{\prod_{i}v_i}$ is bounded.

Any ideas?

anycontinuous function; you can do that to destroy the concavity. Since $\prod_iv_i$ is continuous and positive on $K$, the resulting ratio will still be continuous, hence bounded, just because of the compactness. $\endgroup$ – Alex Degtyarev Jan 21 '15 at 10:52