Let $A=[a\_{ij}]$ be a $3\times 3$ matrix, where $a\_{ii}$ is a real number, and $a\_{ji}=\overline{a\_{ij}}$ is the complex conjugate of $a\_{ij}$ for all $1\leq i,j\leq 3$, i.e $A^t=[\overline{a\_{ij}}]$. Let $\lambda\_1,\lambda\_2,\lambda\_3$ be eigenvalues of $A$ (not necessarily distinct) and $u\_i=(a\_i,b\_i,c\_i)$ be an eigenvector of $A$ corresponding to $\lambda\_i$, $i=1,2,3$. Is this true that
$\sum\_{i=1}^3\frac{|x\_i\overline{y\_i}|}{||u\_i||^2}\leq 1$, where $x\_i,y\_i\in\{a\_i,b\_i,c\_i\}$ (not necessarily distinct)? Is this true in general for an $n\times n$ matrix by replacing $n$ with $3$?



An inequality relating to entries of eigenvectors
The inequality holds for any $n$ if all eigenvalues of the Hermitian matrix $A$ are distinct, so that the eigenvectors form a unitary matrix.$^\ast$


It does not hold if some eigenvalues are identical.  

In that case the corresponding eigenvectors need not be orthogonal, you could choose them nearly parallel, say $u\_1=(1,0,0)$ and $u\_2=(\sqrt{1-\epsilon},\sqrt\epsilon,0)$, with $u\_3=(0,0,1)$; then take $x\_i=a\_i$, $y\_i=a\_i$, and you find that the inequality is violated:
$$\sum\_{i=1}^3\frac{|x\_i\overline{y\_i}|}{||u\_i||^2}=2-\epsilon>1.$$




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$^\ast$ If the eigenvectors $u\_i$ are the rows of an $n\times n$ unitary matrix $U$, the inequality in the OP is the statement that
$$\sum\_{i=1}^n |U\_{ij}\bar{U}\_{ik}|\leq 1,$$
for any pair of integers $j,k\in\{1,2,\ldots n\}$. Define $U'\_{ij}=U\_{ij}\,\exp\bigl(-i\arg[U\_{ij}\bar{U}\_{ik}]\bigr)$,
then
$$\left(\sum\_{i=1}^n |U\_{ij}\bar{U}\_{ik}|\right)^2=\left(\sum\_{i=1}^n U'\_{ij}\bar{U}\_{ik}\right)^2\leq\left(\sum\_{i=1}^n|U'\_{ij}|^2\right)\left(\sum\_{i=1}^n|U\_{ij}|^2\right)=1.$$


