= Solution
Taking the <natural logarithm> of the positive product constraint gives $h(x)=\sum_i a_i\log x_i=0$. Its <gradient> has entries $a_i/x_i>0$, so the equality <constraint qualification> holds everywhere on the <feasible set>. Use the <optimization Lagrangian>
$$
L(x,\lambda)=\sum_i a_ix_i-\lambda\sum_i a_i\log x_i.
$$
Stationarity requires $a_i-\lambda a_i/x_i=0$, so every $x_i=\lambda$. Substitution into the constraint, using $\sum_i a_i=1$, gives $\lambda=1$. Thus the only candidate for a <local minimum> is $x=(1,\ldots,1)$.
To prove global minimality without assuming the inequality to be established, set $u_i=\log x_i$. The constraint becomes the <affine hyperplane> $\sum_i a_iu_i=0$, and the <objective function> becomes $F(u)=\sum_i a_ie^{u_i}$. Its <Hessian matrix> $\operatorname{diag}(a_ie^{u_i})$ is <positive-definite>, so $F$ is <strictly convex>. The elementary tangent inequality $e^t\geq1+t$, with equality only at zero, gives
$$
F(u)\geq\sum_i a_i(1+u_i)=1.
$$
Equality forces every $u_i=0$. Therefore
$$
\boxed{x_*=(1,\ldots,1),\qquad\min\sum_i a_ix_i=1}.
$$
This point is the unique <global minimizer> and the only <local minimizer>. The <logarithm> change of variables also explains why a multiplicative constraint leads to a <convex optimization> argument.
Back to article page
