Let $V$ be a finite dimensional vector space. If $A=\{l_i\}_{i=1}^n \subseteq V^*$ are linear functional such that $\displaystyle\max_{i=1,\dots, n}\{l_i(x)\}\geq 0$ for all $x\in V$, then $A$ is linearly dependent.
A geometrical way to interpret $\displaystyle \max_{i=1,\dots,n}\{l_i(x)\}\geq 0$ for all $x\in V$ is to say that the family of semispaces $\{x\in V\mid l_i(x)\geq 0\}$ cover $V$.
How can I prove this?
Let $f(x) = \max_k l_k(x)$. Note that $f(0) = 0$ and $f(x) \ge 0$ for all $x$. Since $f$ is convex, we have $0 \in \partial f(0)$.
We have $\partial f(0) = \operatorname{co} \{ l_k \}_{k \in I(0)} $, where $I(x) = \{ k | f(x) = l_k(x) \}$.
In particular, there is a (non zero) convex multiplier $\mu$ such that $\sum_k \mu_k l_k = 0$ from which it follows that $A$ is not linearly independent.