= Solution
Let $\rho_j=|\alpha_j\rangle\langle\alpha_j|$. For a two-outcome measurement with effect $0\leq M\leq I$ interpreted as guess zero,
$$
P_s=\frac12\operatorname{tr}(M\rho_0)
\frac12\operatorname{tr}((I-M)\rho_1)
=\frac12+\frac12\operatorname{tr}[M(\rho_0-\rho_1)].
$$
The positive-eigenspace projector of $\rho_0-\rho_1$ maximizes the last trace, giving the Helstrom formula
$$
P_s^{\rm opt}=\frac12+\frac14\|\rho_0-\rho_1\|_1.
$$
On the span of the two states, $\rho_0-\rho_1$ has eigenvalues
$$
\mathord\pm\sqrt{1-|\langle\alpha_0|\alpha_1\rangle|^2}.
$$
Therefore
$$
\boxed{
P_s\leq\frac12\left(1+
\sqrt{1-|\langle\alpha_0|\alpha_1\rangle|^2}\right)},
$$
and the positive/negative eigenspace measurement attains equality.
The right side equals one exactly when $|\langle\alpha_0|\alpha_1\rangle|=0$. Thus two pure states are perfectly distinguishable exactly when they are orthogonal.
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