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# An interesting quadratic form bound.
Let $A$ be any $n\times n$ real symmetric matrix, whose columns are $\vec a_{1},\ldots,\vec a_{n}$, say. Denote $\tilde A$ to be the matrix where $$
\tilde A = \text{diag}(\Vert \vec a_{1}\Vert ,\ldots,\Vert \vec a_{n}\Vert)
$$
We claim the following: For any $x \in \mathbb{R}^{n}$, we have $$
(x,Ax)\le \sqrt{n}(x,\tilde A x)
$$where $(x,y)$ is usual dot product. If $x=\vec 0$ then we are done. So suffices to consider $x\neq \vec 0$.
Let us write $x = (x_{1},\ldots,x_{n})^{T}$. Then $$
Ax = x_{1}\vec a_{1}+x_{2}\vec a_{2}+\cdots + x_{n}\vec a_{n},
$$And so $$
(x,Ax)=x_{1}(x,\vec a_{1})+x_{2}(x,\vec a_{2})+\cdots +x_{n}(x,\vec a_{n})
$$
Now, by Cauchy-Schwarz, $$
|(x,\vec a_{i})| \le \Vert x\Vert \cdot \Vert \vec a_{i}\Vert
$$So, $$
|(x,Ax)|\le \Vert x\Vert (|x_{1}|\cdot \Vert\vec a_{1}\Vert + \cdots +|x_{n}|\cdot \Vert \vec a_{n}\Vert)
$$
Now what is $\tilde A x$? Well, $$
\tilde Ax = \begin{pmatrix}
x_{1}\Vert\vec a_{1}\Vert \\
x_{2}\Vert\vec a_{2}\Vert \\
\cdots \\
x_{n}\Vert\vec a_{n}\Vert
\end{pmatrix}
$$so $$
\begin{align*}
(x,\tilde A x) &= x_{1}^{2}\Vert\vec a_{1}\Vert + \cdots +x_{n}^{2}\Vert a_{n}\Vert \\
&=\Vert x\Vert \left(\frac{x_{1}^{2}}{\Vert x\Vert} \Vert\vec a_{1}\Vert+\cdots +\frac{x_{n}^{2}}{\Vert x\Vert} \Vert\vec a_{n}\Vert\right)
\end{align*}
$$
Hmm. Need Jensen/AMGM sort of concavity bounds or something?
Ok, experimentally we need to use the fact that $A$ is symmetric, otherwise counterexamples exist.