Bounding a quadratic form by eigenvalues

Diagonalization. Since $A$ is symmetric, there is an orthogonal matrix $Q$ with $A=QDQ^T$, $D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$.

Change of variables. Let $Y=Q^TX$. Then $X^TAX=Y^TDY=\sum_i\lambda_i y_i^2$.

Bounding. Since $\lambda_1\le\lambda_i\le\lambda_n$:
$$\lambda_1\sum_i y_i^2\le\sum_i\lambda_i y_i^2\le\lambda_n\sum_i y_i^2.$$
But $\sum_i y_i^2=\|Y\|^2=\|X\|^2$ (since $Q$ is orthogonal).

Conclusion. $\lambda_1\|X\|^2\le X^TAX\le\lambda_n\|X\|^2$.