I try really hard to prove this Question. let $A_{nXn}(\mathbb{R})$ Symmetric matrix $A=A^t$ let $\lambda$ be the greatest Eigenvalue of A. we will define over the field $\mathbb{R}$ with the standard inner product a Linear map such that
$\forall v \in R^n$, $v \to Av$
prove : $$\max_{\displaystyle{\forall v\in R^n}\atop{\displaystyle ||v||=1}} < Av,v >=\lambda$$
Symmetric matrices over ${\mathbb{R}}$ are diagonalizable; choose an orthonormal basis $x_1, \dots, x_n$ of eigenvectors of $A$ belonging to eigenvalues $\lambda_1, \dots, \lambda_n$. (That is, $\left<x_i, x_j\right> = 0$ for $i\not = j$; each $\left<x_i, x_i\right> = 1$; and $Ax_i = \lambda_i x_i$.) Then for $x = t_1 x_1 + \cdots + t_n x_n$, we have $$\left<Ax, x\right> = \sum t_i t_j \left<Ax_i, x_j\right> = t_1^2 \lambda_1 + \cdots + t_n^2 \lambda_n \leq (t_1^2 + \cdots + t_n^2)\lambda = \lambda\left<x, x\right>$$ (provided $\lambda \geq 0)$. The result follows.