Let $\mathfrak g$ be a complex simple Lie algebra. 
Fix a Cartan subalgebra $\mathfrak h$ of $\mathfrak g$, let $\Delta$ denote the corresponding root system. 
Pick a partial order on $\mathfrak h$, which induces a positive root system $\Delta^+$, fundamental Weyl chamber, etc. 


I wish a reference for the following result, which must be well known. 
My proof is case-by-case and a bit long. 



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> For all but finitely many **finite dimensional** irreducible representations $(\pi, V)$ of $\mathfrak g$, there is a weight $\mu$ of $\pi$ lying in the fundamental Weyl chamber. 
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Recall that $\mu\in\mathfrak h^\*$ is called a weight of $\pi$ if $V(\mu):=\{v\in V: \pi(X)\cdot v=\mu(X)\, v\quad\forall \, X\in\mathfrak h\}$ is non-zero. 
Furthermore, the fundamental Weyl chamber is given by elements $\mu\in\mathfrak h^\*$ satisfying $\langle \mu,\alpha\rangle >0$ for all $\alpha\in\Delta^+$. 


The proof is immediately reduced to consider irreducible representations of $\mathfrak g$ with highest weight lying in a face of the fundamental Weyl chamber. 
Furthermore, it reduces to check irreducible representations with highest weight a multiple of a fundamental weight. 


It is easy to see that the result is no longer valid when $\mathfrak g$ is assumed semisimple in place of simple. 



Existence of a weight of a representation in the fundamental Weyl chamber
Here's maybe another (more conceptual?) way to think about it.


First of all, if $\mu\_1$ is a dominant weight which appears with nonzero multiplicty in $V^{\lambda\_1}$, and $\mu\_2$ is a dominant weight which appears with nonzero multiplicty in $V^{\lambda\_2}$, then $\mu\_1+\mu\_2$ is a dominant weight which appears with nonzero multiplicity in $V^{\lambda\_1+\lambda\_2}$. (This is just saying that if $\lambda\_1-\mu\_1$ is a nonnegative sum of simple roots, and ditto for $\lambda\_2-\mu\_2$, then definitely $(\lambda\_1+\lambda\_2)-(\mu\_1-\mu\_2)$ is as well.)


The previous paragraph reduces the problem to showing that if $\lambda=\omega\_i$ is a fundamental weight, then for all $m\gg0$ sufficiently large, $V^{m\omega\_i}$ contains a weight strictly in the fundamental chamber. Let's show this. 


Recall the *Weyl vector* $\rho:=\frac{1}{2}\sum\_{\alpha\in\Phi^+}\alpha$, which is well known to satisfy $\rho= \sum\_{j=1}^{n}\omega\_j$. Let $\Phi\_i$ denote the maximal parabolic root system generated by the simple roots $\alpha\_j$ for $j\neq i$, and let's use the notation $\rho\_i :=\frac{1}{2}\sum\_{\alpha\in\Phi\_i^+}\alpha$ for its "parabolic Weyl vector." Note that $\langle \rho\_i,\alpha^\vee\_j\rangle = 1$ for all $j\neq i$ precisely because this is the Weyl vector of the parabolic root system, while $\langle \rho\_i, \alpha^\vee\_i \rangle < 0$ since no factor of $\alpha\_i$ appears at all in $\rho\_i$ but some $\alpha\_j$ for $j$ a node adjacent to $i$ in the Dynkin diagram does appear.


Therefore $2\rho - 2\rho\_i = \sum\_{\alpha\in \Phi^+\setminus \Phi\_i^+}\alpha= c\omega\_i$ for some integer $c > 0$. But one of the $\alpha \in \Phi^+\setminus \Phi\_i^+$ is the *highest root* $\theta$ of $\Phi$, which when written $\theta=\sum\_{j=1}^{n}a\_j\alpha\_j$ satisfies $a\_j \geq 1$. So there is some multiple $k$ of $2\rho - 2\rho\_i$ for which $k(2\rho - 2\rho\_i) - 2\rho$ is a nonnegative sum of simple roots. This is to say the strictly dominant weight $2\rho$ appears in $V^{kc\omega\_i}$. And then for any $m\geq kc$, the strictly dominant weight $2\rho + (m-kc)\omega\_i$ appears in $V^{m\omega\_i}$, completing the proof.


