= Solution
The equality-constrained <Lagrangian sufficiency theorem> says: if $x^*$ is feasible and, for some fixed multipliers, it globally minimizes the <optimization Lagrangian> over all $x$, then it globally minimizes the objective over feasible $x$. On the feasible set the constraint terms vanish, so the two minimizations agree. No convexity assumption is needed for this version once global minimization of the <optimization Lagrangian> has been established.
Let $S_1=\sum_i a_i$, $S_2=\sum_i a_i^2$, and $\Delta=nS_2-S_1^2$. Since
$$
\Delta=n\sum_i(a_i-S_1/n)^2>0,
$$
the two constraints are independent. For the squared <Euclidean norm> objective in the original PDF, take
$$
L(x,\lambda,\mu)=\sum_i x_i^2-2\lambda\left(\sum_i x_i-1\right)-2\mu\sum_i a_ix_i.
$$
<Completing the square> shows that its unique global minimizer has $x_i=\lambda+\mu a_i$. The constraints become $n\lambda+S_1\mu=1$, $S_1\lambda+S_2\mu=0$, giving $\lambda=S_2/\Delta$, $\mu=-S_1/\Delta$. The <Lagrangian sufficiency theorem> therefore gives
$$
\boxed{x_i^*=\frac{S_2-S_1a_i}{nS_2-S_1^2},\qquad \min\sum_i x_i^2=\frac{S_2}{nS_2-S_1^2}.}
$$
For the value, use $\sum_i(x_i^*)^2=\lambda\sum_i x_i^*+\mu\sum_i a_ix_i^*=\lambda$. Uniqueness also follows directly: for every feasible $x$, the cross term in $\sum_i(x_i-x_i^*)^2$ vanishes, so the difference of objectives equals that strictly positive sum unless $x=x^*$. This is the <minimum squared norm under two affine constraints>.
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