 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 $E_{i}v=0$ and $H_{i}v=\lambda (H_i)v$.

Since $F_i$ decreases the weight by ${\alpha }_i$ and there are only finitely many weights of a representation, let $n\ge 0$ be maximal with $F_i^{n}v\ne 0$. Induction using the two bracket relations gives the sl2 highest-weight lowering formulaApplying this at $k=n+1$, the left side vanishes, and the nonzero vector $F_i^{n}v$ on the right forces $\lambda (H_i)=n$. This holds for every simple root, soThus the highest weight is a dominant integral weight, in particular an element of the Closed dominant Weyl chamber.