In Lagrange Multiplier theorem.
Consider function $f(\vec{x}):R^n\rightarrow R$ that need to be optimized for some $\vec{x}$. And constraint $g(\vec{x})=0$| given by function $g(\vec{x}):R^n\rightarrow R$ that need to be satisfied.
On infinitesimal scale if $\vec{y}=\vec{x}+\Delta\vec{x}$. $$\begin{align} g(\vec{y})-g(\vec{x})=\triangledown g(\vec{x})\Delta\vec{x}\newline \triangledown g(\vec{x})\Delta\vec{x}\stackrel{(a)}{=}0\newline g(\vec{x})=0\newline g(\vec{x}+\Delta \vec{x})=0 \end{align}$$
Now assuming that $\vec{x}$ is minimum of constrained problem. Then any displacement from $\vec{x}$ to $\vec{x}+\vec{v}$ still satisfying constraint, should increase $f(\vec{x})$ to $f(\vec{x}+\vec{v})$. On infinitesimal scale only consider displacements $\vec{v}$ preserving $g(\vec{x}+\vec{v})=0$. From equality (a) it means that $\triangledown g(\vec{x})\Delta\vec{v}=0$. And $\triangledown f(\vec{x})\vec{u}\geq 0$, otherwise $f(\vec{x})$ is not minimized and $\vec{x}$ is not solution of problem.
As I understand $\nabla f(\vec{x})\vec{u}$ should be greater or equal to 0, but why every proof I see for example states that $\nabla f(\vec{x})\vec{u}=0$. And thereof claims that $\nabla g$ and $\nabla f$ are parallel.
