Here's a generalization of the result that eigenvectors of a real symmetric matrix with distinct eigenvalues are orthogonal:

Let $x$ and $y$ be unit eigenvalues of a (not necessarily symmetric) matrix $A$ with eigenvalues $\lambda$ and $\mu$, respectively. Then $$ \lambda\langle x,y\rangle =\langle Ax,y\rangle =\langle x,A^\mathrm{T}y\rangle =\langle x,Ay\rangle-\langle x,(A-A^\mathrm{T})y\rangle =\overline{\mu}\langle x,y\rangle-\langle x,(A-A^\mathrm{T})y\rangle. $$ Rearranging and applying standard inequalities then yields $$ |\langle x,y\rangle| \leq\frac{\|A-A^\mathrm{T}\|_2}{|\lambda-\overline{\mu}|}. $$ The numerator of this fraction makes intuitive sense, since any correlation between eigenvectors must come from the skew-symmetric part of the matrix. Furthermore, the denominator provides the relationship you suggested in your question, especially when the eigenvalues are real.

Here's a generalization of the result that eigenvectors of a real symmetric matrix with distinct eigenvalues are orthogonal:

Let $x$ and $y$ be unit eigenvalues of a real matrix $A$ with eigenvalues $\lambda$ and $\mu$, respectively. Then $$ \lambda\langle x,y\rangle =\langle Ax,y\rangle =\langle x,A^\mathrm{T}y\rangle =\langle x,Ay\rangle-\langle x,(A-A^\mathrm{T})y\rangle =\overline{\mu}\langle x,y\rangle-\langle x,(A-A^\mathrm{T})y\rangle. $$ Rearranging and applying standard inequalities then yields $$ |\langle x,y\rangle| \leq\frac{\|A-A^\mathrm{T}\|_2}{|\lambda-\overline{\mu}|}. $$ The numerator of this fraction makes intuitive sense, since any correlation between eigenvectors must come from the skew-symmetric part of the matrix. Furthermore, the denominator provides the relationship you suggested in your question, especially when the eigenvalues are real.