This is just the details of the first step of Alexander Eremenko's answer (so upvote his answer if you like mine), which I think is by far the most elementary. You only need two facts: A continuous function on a compact set in $R^n$ achieves its maximum (or minimum), and the derivative of a smooth function vanishes at a local maximum. And there's no need for Lagrange multipliers at all.

Let $C$ be any closed annulus centered at $0$. The function $$ R(x) = \frac{x\cdot Ax}{x\cdot x}, $$ is continuous on $R^n\backslash\{0\}$ and therefore achieves a maximum on $C$. Since $R$ is homogeneous of degree $0$, any maximum point $x \in C$ is a maximum point on all of $R^n\backslash\{0\}$. Therefore, for any $v \in R^n$, $t = 0$ is a local maximum for the function $$ f(t) = R(x + tv). $$ Differentiating this, we get $$ 0 = f'(0) = \frac{2}{x\cdot x}[Ax - R(x) x]\cdot v $$ This holds for any $v$ and therefore $x$ is an eigenvector of $A$ with eigenvalue $R(x)$.