I'm attempting to figure out why Lagrange Multipliers work, and I have the following setup. Suppose we wish to maximize $F : \mathbb{R}^n\to\mathbb{R}$ with respect to the constraint $G = c$ where $G:\mathbb{R}^n\to\mathbb{R}$. Let $L$ be the level surface $G = c$. Suppose $x_0\in L$, and $u$ is tangent to $L$ at $x_0$. I want to show that if $\nabla F\cdot u > 0$ then we can move along $L$ and find a point $x_0'\in L$ with $F(x_0')>F(x_0)$, hence requiring that $\nabla F\cdot u = 0$ for $x_0$ to be a maximum.
My problem is that the condition $f'(x_0)>0$ isn't enough to guarantee that there is an interval $[x_0,\ x_0+\delta]$ on which $f$ is increasing (for example, take $f(x) = x^2\sin^2\left(\frac1x\right) + \frac12 x$ for nonzero $x$, and $f(0)=0$). So essentially, if we look at $f(t) = F(x_0+tu)$, the statement $f'(0) > 0$ is not enough to assume that $f$ is strictly increasing on any neighbourhood of zero, and so $F$ isn't increasing in any neighbourhood of $x_0$ meaning we can't necessarily guarantee that it's increasing along $L$ in the direction of $u$. Hence, $\nabla F\cdot u>0$ isn't enough to guarantee that we can find another point $x_0'\in L$ near $x_0$ so that $F(x_0')>F(x_0)$.
If my logic isn't flawed, then how come Lagrange multipliers work? How do we know that $\nabla F\cdot u = 0$ for all tangent vectors $u$ to $L$ at $x_0$ is enough to guarantee that $x_0$ maximizes $F$ on $L$? Is it like, $\nabla F\cdot u>0$ isn't enough to guarantee that $x_0$ isn't a maximum, but $\nabla F\cdot u = 0$ is?