6. Let $P$ be a fixed point on the circle $x^{2}+y^{2}=1$, and points $Q$ and $R$ are inside or on the circumference of the circle, and $\triangle P Q R$ is an equilateral triangle with side length $\frac{2}{\sqrt{3}}$. Then the minimum value of $|O Q|^{2}+|O R|^{2}$ is
6. $\frac{2}{3}$.

Without loss of generality, let $P(-1,0)$, then the equation of the circle with $P$ as the center and $\frac{2}{\sqrt{3}}$ as the radius is
$$
(x+1)^{2}+y^{2}=\frac{4}{3} \text {. }
$$

It is easy to see that the two intersection points of $\odot P$ and the circle $x^{2}+y^{2}=1$ are $A\left(-\frac{1}{3}, \frac{2 \sqrt{2}}{3}\right)$ and $B\left(-\frac{1}{3},-\frac{2 \sqrt{2}}{3}\right)$. Then the vertices $Q$ and $R$ of the equilateral $\triangle PQR$ lie on the minor arc $\overparen{AB}$ of $\odot P$, and $\angle QPR=\frac{\pi}{3}$.
Let $Q\left(-1+\frac{2}{\sqrt{3}} \cos \theta, \frac{2}{\sqrt{3}} \sin \theta\right)$, then
$$
R\left(-1+\frac{2}{\sqrt{3}} \cos \left(\theta+\frac{\pi}{3}\right), \frac{2}{\sqrt{3}} \sin \left(\theta+\frac{\pi}{3}\right)\right) \text {. }
$$

It is easy to see that $-\arctan \sqrt{2} \leqslant \theta \leqslant \arctan \sqrt{2}$. Thus,
$$
\begin{array}{l}
|OQ|^{2}+|OR|^{2} \\
=\left(-1+\frac{2}{\sqrt{3}} \cos \theta\right)^{2}+\left(\frac{2}{\sqrt{3}} \sin \theta\right)^{2}+ \\
\quad\left[-1+\frac{2}{\sqrt{3}} \cos \left(\theta+\frac{\pi}{3}\right)\right]^{2}+\left[\frac{2}{\sqrt{3}} \sin \left(\theta+\frac{\pi}{3}\right)\right]^{2} \\
=\frac{14}{3}-\frac{4}{\sqrt{3}}\left[\cos \theta+\cos \left(\theta+\frac{\pi}{3}\right)\right] \\
=\frac{14}{3}-\frac{8}{\sqrt{3}} \cos \left(\theta+\frac{\pi}{6}\right) \cdot \cos \frac{\pi}{6} \\
=\frac{14}{3}-4 \cos \left(\theta+\frac{\pi}{6}\right) .
\end{array}
$$

When $\theta=-\frac{\pi}{6}$,
$\left(|OQ|^{2}+|OR|^{2}\right)_{\min }=\frac{14}{3}-4=\frac{2}{3}$.
Editor's note: The minimum value can also be obtained by the median formula of a triangle.