Let us suppose we have a quadratic form
$$
Q(\mathbf{x})=\sum_{i,j=1}^n A_{ij}x_ix_j
$$
where $A$ is a real symmetric square matrix and $\mathbf{x}\in\mathbb{R}^n$, and let us suppose that $Q$ is positive definite (or semi-definite).
Let's now call $Q^*$ the quadratic form associated to the matrix $A^2$, i.e.
$$
Q^*(\mathbf{x})=\sum_{i,j=1}^n\left(\sum_{k=1}^n A_{ik}A_{kj}\right)x_ix_j
$$
Is it true that
$$
Q^*(\mathbf{x})\leq\lambda_\text{max}Q(\mathbf{x})
$$
where $\lambda_{\text{max}}$ is the maximum positive eigevalue of $Q$?
My tentative proof is that the value of a quadratic form should be independent of the choice of an orthonormal basis in $\mathbb{R}^n$, and if we choose the basis in which the matrix $A$ is diagonal, the result seems rather evident:
$$
Q^*(\mathbf{x})=\lambda_1^2x_1^2+\ldots+\lambda_n^2x_n^2\leq\lambda_\text{max}(\lambda_1x_1^2+\ldots+\lambda_2x_2^2)=\lambda_\text{max}Q(\mathbf{x}).
$$
Best Answer
Your intuition is fine. Here is the formal way:
Define $A=U^TDU$ as the eigendecomposition of $A$, with $\det(U) = 1$ and $U^{-1} = U^T$ since $A$ is symmetric. Note that the diagonal entries of diagonal matrix $D$ will be positive (for $A$ positive definite) and will be the eigenvalues $\lambda_i$ of $A$. Let $u=\frac{x}{\sqrt{x^Tx}}$ and $z=Uu$. Note that $u$ and $z$ are unit norm vectors. Then \begin{align} \frac{Q(x)}{||x||^2}=\frac{x^TAx}{x^{T}x}=u^{T}Au=u^{T}U^{T}DUu=z^TDz=\sum_{i=1}^{N}\lambda_i|z_i|^2 \end{align} where $z_i$'s are entries of the vector $z$. With the same method, \begin{align} \frac{Q^*(x)}{||x||^2}=\frac{x^TA^2x}{x^{T}x}=u^{T}A^2u=u^{T}U^{T}DUU^{T}DUu \end{align} Since $U^{-1} = U^T$, we have $U U^T = \bf 1$ , where $\bf 1$ is the unit matrix. So \begin{align} \frac{Q^*(x)}{||x||^2}=u^{T}U^{T}D^2Uu=z^TD^2z=\sum_{i=1}^{N}\lambda_i^2|z_i|^2 \end{align} and finally \begin{align} Q^*(x)= ||x||^2 \sum_{i=1}^{N}\lambda_i^2|z_i|^2 \le ||x||^2 \lambda_ {max} \sum_{i=1}^{N}\lambda_i|z_i|^2 = \lambda_ {max} Q(x) \end{align} which establishes the claim.