= Solution
Let $v$ be a <highest-weight vector> of weight $\lambda$. For a <simple root> $\alpha_i$, choose its <sl2 subalgebra associated with a root>, with operators $E_i,F_i,H_i$ satisfying $[E_i,F_i]=H_i$, $[H_i,F_i]=-2F_i$, and $H_i=\alpha_i^\vee$. We have $E_iv=0$ and $H_iv=\lambda(H_i)v$.
Since $F_i$ decreases the weight by $\alpha_i$ and there are only finitely many <weights of a representation>, let $n\geq0$ be maximal with $F_i^nv\ne0$. Induction using the two bracket relations gives the <sl2 highest-weight lowering formula>
$$
E_iF_i^kv=k\bigl(\lambda(H_i)-k+1\bigr)F_i^{k-1}v.
$$
Applying this at $k=n+1$, the left side vanishes, and the nonzero vector $F_i^nv$ on the right forces $\lambda(H_i)=n$. This holds for every <simple root>, so
$$
\boxed{\langle\lambda,\alpha_i^\vee\rangle\in\mathbb Z_{\geq0}\text{ for all }i.}
$$
Thus the <highest weight> is a <dominant integral weight>, in particular an element of the <Closed dominant Weyl chamber>.
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