How to prove this? The existence of solutions to linear inequalities A system of real homogeneous linear inequalities $\lambda_i>0$, $i=1,2,\ldots,m$, has a solution if and only if there is no nontrivial linear dependence with nonnegative coefficients among the $\lambda_i$. For example, $\lambda_i=\sum a_{ij}x_j$.
 A: Here is a quick argument (I had to fill in the details recently). 

Theorem 
  Let $v_1, \dots, v_m \in \mathbf{R}^n$. Then, there exists $x \in \mathbf{R}^n$ such that $v_i'x > 0$ for all $i = 1, \dots, m$ if and only if the relation $\sum_{i=1}^m \mu_i v_i = 0$ for some $\mu_1, \dots, \mu_m \in [0,\infty)$ holds, then, $\mu_j = 0$ for $j=1, \dots, m$. 

Proof. $(\Rightarrow)$ is obvious. 
$(\Leftarrow)$ Consider the convex hull $C$ of $\{v_i\}$. Then, $C$ is compact (Why? See below for a quick sketch of the proof). Thus, the real valued function $v \mapsto \|v\|$ on $C$ attains its minimum on $C$. Say $x := x_{\min}$ is the minimizer. Then, $\|x\| > 0$ (by hypothesis, $0 \notin C$).
Then, one verifies directly that $v'x > 0$ for all $v \in C$: this is clear for $v =x$; now, if $v \neq x$, by convexity of $C$, we have $tv+(1-t)x \in C$ for all $t \in [0,1]$ whence $\|tv+(1-t)x\| \geq \|x\|$ for all $t \in [0,1]$; it now follows that 
$$t^2\|v-x\|^2+2t\langle v-x, x\rangle \geqslant 0 \text{ for all } t \in [0,1].$$ This suggests that we look at the function 
$$f(t) = t^2\|v-x\|^2+2t\langle v-x, x\rangle,\quad t \in \mathbf{R}$$
for further analysis. The graph of $f$ is a parabola open upwards and for $t \in [0,1]$, $f(t) \geq 0$. Now, setting derivative equal to $0$, we obtain $$t_0=-\frac{\langle v-x, x\rangle}{\|v-x\|^2}$$
where $f$ attains the minimum. Clearly, $t_0 \leq 0$ (try sketching the graph of $f$ reminding yourself that $f(t) \geq 0$ for all $t \in [0,1]$!). In particular, 
$$\langle v,x\rangle \geq \langle x, x \rangle > 0.$$   
Thus, in particular, we have found an $x \in\mathbf{R}^n$ such that $v_i'x > 0$. 
Compactness of Convex hull of compact sets The key idea is that the standard simplex is compact. Thus, one takes a sequence from $C$, then, writing it as convex linear combinations of $\{v_i\}$, we get a sequence in the simplex. By compactness of the simplex, we obtain a convergent subsequence, which in turn yields a subsequence of the original sequence, which converges! 
Remark. This result is really another avatar of hyperplane separation theorem for convex sets: given two disjoint convex sets, there is an hyperplane separating them. (A weaker version of this might do the job too!)
