Find a global minimum for $f(x_1,\dots,x_n)=\sum_{i=1}^n \frac{1}{x_i}$ under contains $\sum_{i=1}^n x_ia_i^2=1$ for $x_i, a_i > 0$ $(\forall 1 \leq i \leq n)$.
Applying Lagrange multiplier theorem we get one point: $\frac{1}{a_1\sum_{i=1}^n a_i}, \dots, \frac{1}{a_n\sum_{i=1}^n a_i}$. We can show that $f \rightarrow\infty$ when $x \rightarrow 0$, so the global minimum isn't near zero. Intuition says that there are no more options for a global minimum, but I'm not sure how to prove it.
The objective function is convex because it is a sum of convex functions (for $x_i > 0$). The constraint is affine. Hence the problem satisfies the assumptions of a convex minimisation problem: the global minimum is unique.