6. Given real numbers $a, b$ satisfy $a^{2} \geqslant 8 b$. Then
$$
(1-a)^{2}+(1-2 b)^{2}+(a-2 b)^{2}
$$

the minimum value is $\qquad$
6. $\frac{9}{8}$.

For $a^{2} \geqslant 8 b$, we can use the Vieta substitution, and without loss of generality, assume
$$
\begin{array}{l}
a=x+y, 2 b=x y . \\
\text { Then } (1-a)^{2}+(1-2 b)^{2}+(a-2 b)^{2} \\
\left.=2\left((x+y)^{2}+x^{2} y^{2}-x y(x+y)-(x+y)+1\right)\right) \\
=2\left(x^{2}-x+1\right)\left(y^{2}-y+1\right) \geqslant \frac{9}{8},
\end{array}
$$

Equality holds if and only if $x=y=\frac{1}{2}$, in which case,
$$
a=1, b=\frac{1}{8} \text {. }
$$