Let $f(\mathbf{x})$ and $g_i(\mathbf{x}), i=0,{\ldots},m$ be differentiable scalar functions; $\mathbf{x} \in R^n$

We will find local extremes of the function $f(\mathbf{x})$ where $\nabla f=0$ This can be proved by contradiction: $$ \nabla f \neq 0 $$ $$ \Leftrightarrow \exists \epsilon_0 > 0, \forall \epsilon; 0<\epsilon<\epsilon_0: f(\mathbf{x}-\epsilon \nabla f) < f(\mathbf{x}) < f(\mathbf{x+\epsilon \nabla f}) $$ but then $f(\mathbf{x})$ is not a local extreme.

Now we put up some conditions, such that we should find the $\mathbf{x} \in S \subset R^n$ that gives a local extreme of $f$ Let $S=\bigcap_{i=1}^m S_i$ and let $S_i$ be defined so that $g_i(\mathbf{x})=0 \forall \mathbf{x} \in S_i$

Any vector $\mathbf{x} \in R^n$ can have one component perpendicular to the subset $S_i$ (for visualization, think $n=3$ and let $S_i$ be a flat surface). $\nabla g_i$ will be perpendicular to $S_i$ because: $$ \exists \epsilon_0>0, \forall \epsilon; 0<\epsilon<\epsilon_0: g_i(\mathbf{x}-\epsilon \nabla g_i)<g_i(\mathbf{x})< g_i(\mathbf{x}+\epsilon \nabla g_i) $$ But $g_i(\mathbf{x})=0$ so any vector $\mathbf{x}+\epsilon \nabla g_i$ must be outside $S_i$ and also outside $S$ (todo: I have proved that there might exist a component perpendicular to each subset $S_i$ but not that there exists only one; this should be done)

By the argument above, $\nabla f$ must be zero - but now we can ignore all components of $\nabla f$ perpendicular to $S$ (todo: this should be expressed more formally and proved)

So we will have a local extreme within $S_i$ if there exists a $\lambda_i$ such that $$ \nabla f = \lambda_i \nabla g_i $$

We will have local extreme(s) within $S$ where there exists a set $\lambda_i, i=1,{\ldots},m$ such that $$ \nabla f = \sum \lambda_i \nabla g_i $$