I am trying to prove the following statement: for $A$ a $n\times n$ symmetric matrix (with real coefficients) such that $\text{tr}(A)=0$, one has for every vector $e$ of euclidean norm 1 the following inequality $$ |Ae - (Ae,e)e| \le \frac{1}{\sqrt{2}}\left\| A \right\|_F, $$ where $|\cdot|$ denotes the usual euclidean norm on vectors (and $(\cdot,\cdot)$ is the associated scalar product) and $\left\| \cdot \right\|_F$ is the Frobenius norm on matrices, associated to the scalar product $$ (B,C)_F = \sum_{i=1}^n \sum_{j=1}^n b_{i,j}c_{i,j}, $$ if $B = (b_{i,j})_{1\le i,j\le n}$ and $C = (c_{i,j})_{1\le i,j\le n}$.
Here are several facts that I have collected: one can easily get the constant $1$ instead of $\frac{1}{\sqrt{2}}$, the fact $A$ is trace free is important (for example, the inequality is wrong when $A= \lambda I$, where $I$ is the identity matrix and $\lambda \neq 0$), the symmetry seems to play a role too.
I have not been able to find a proof of this inequality (I found it without proof in a book).
This is true for all real symmetric matrices, not just for traceless ones. Since Euclidean norm, Euclidean inner product and Frobenius norm are unitarily invariant, by a change of orthonormal basis, we may assume that $u=e_n=(0,\ldots,0,1)^T$. Let the last column of $A$ be $\pmatrix{\mathbf a\\ a_{nn}}$, where $\mathbf a\in\mathbb R^{n-1}$. Then the inequality reduces to $\|\mathbf a\|_2\le\frac1{\sqrt{2}}\|A\|_F$ and in turn $\|\mathbf a\|_2^2\le\frac12\|A\|_F^2$, which is obviously true.