= Solution
Use the <coroot> $\alpha^\vee$ in the pairing. The original PDF retains this mark, which is lost in the local TeX transcription. Fix any <positive root> $\alpha$, and put $n=\langle\lambda,\alpha^\vee\rangle$. For a <dominant integral weight>, $n$ is a nonnegative integer; positive <coroots> are nonnegative integral combinations of simple <coroots>.
Let $v_\lambda$ be a <highest-weight vector>, and restrict to the <sl2 subalgebra associated with a root> $\alpha$. It is killed by its <raising operator> $e_\alpha$ and has $h_\alpha$ eigenvalue $n$. If $n\geq1$, its <lowering operator> produces a nonzero vector: otherwise $[e_\alpha,f_\alpha]v_\lambda=n v_\lambda$ would vanish. Therefore $\lambda-\alpha$ is a weight.
Choose a <Weyl group> invariant positive-definite <inner product> on the real weight space. All weights of a <minuscule representation> have the same norm, since they belong to $W\lambda$. But
$$
\|\lambda-\alpha\|^2-\|\lambda\|^2
=-2(\lambda,\alpha)+(\alpha,\alpha)
=(1-n)(\alpha,\alpha).
$$
If $n\geq2$, this is negative, a contradiction. Thus for every <positive root>,
$$
\boxed{\langle\lambda,\alpha^\vee\rangle\in\{0,1\},\quad\text{in particular }\langle\lambda,\alpha^\vee\rangle\leq1.}
$$
When $n=1$, the lower weight is precisely the <Weyl reflection> $s_\alpha\lambda=\lambda-\alpha$, consistent with the orbit condition.
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