Let $f,g\in C^1(U,\mathbb{R})$, such that $U$ is open and non-empty, and let $a\in U$ be a value such that $f$ attains a local extreum under the constraint $g(x)=0$ and $\nabla g(a)\neq 0$. Then there is $\lambda\in \mathbb{R}$, s.t. $$\nabla f(a)=\lambda \nabla g(a)$$
I know how to prove the above Lagrange multiplier theorem, but was wondering how to generalize it to $n$ constraints. That is: $$\nabla f(a)=\sum_{i=1}^n \lambda_i (\nabla g_i)(a)$$
I know the proof is similar, but I am not sure how to formalize it (using the implicit function theorem and not manifolds). Also, are there any good links for a proof to the above generalization. I could only find one, and couldn't understand it.
Any help would be much appreciated. Thanks in advance!
Consider $\textbf {g}(\textbf {x})=\textbf{0}$ as a manifold (Namely, $\textbf{G}$) (this can be done under regularity conditions, such as $\textbf{g}\in C^{\infty}$)
From this point of view, the question of looking for the necessary condition for a point to minimize $f_{|\textbf{g}=\textbf{0}}$ can be restated as follows: "given $\varphi(t): I\rightarrow\textbf{G},\ \varphi(0)=\textbf{x*}$, which are the necessary condition for $\textbf{x*}$ to be a minimum point?"
From a straight application of Fermat's theorem, we get that, $\forall \varphi: \varphi(0)=\textbf{x*}$, $\frac{\partial}{\partial t}f(\varphi(t))|_{t=0}=0$.
Thus, using the chain rule: $\nabla f(\textbf{x*})\cdot \varphi'(0)=0$
This means that $\nabla f(\textbf{x*})\in T_{\textbf{x*}}\textbf{G}^{\perp}$, and so it can be expressed as a linear combination of $\nabla g_m(\textbf{x*})$
$\begin{equation} \nabla f(\textbf{x*})=\sum_m\lambda_m\cdot\nabla g_m(\textbf{x*}) \end{equation}$