6. Planar vectors $\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}$ satisfy $|\boldsymbol{a}|=1, \boldsymbol{a} \cdot \boldsymbol{b}=\boldsymbol{b} \cdot \boldsymbol{c}=1,|\boldsymbol{a}-\boldsymbol{b}+\boldsymbol{c}| \leqslant 2 \sqrt{2}$, then the maximum value of $\boldsymbol{a} \cdot \boldsymbol{c}$ is $\qquad$
Let $\boldsymbol{a}=(1,0)$, according to the problem, $\boldsymbol{b}=(1, t), \boldsymbol{c}=(x, y) \Rightarrow x+t y=1$, take $x=s, y=\frac{1-s}{t} \Rightarrow \boldsymbol{c}=\left(s, \frac{1-s}{t}\right)$. Since $\boldsymbol{a} \cdot \boldsymbol{c}=s$, let's assume $s>1$, then $\boldsymbol{a}-\boldsymbol{b}+\boldsymbol{c}=\left(s,-t+\frac{1-s}{t}\right),|\boldsymbol{a}-\boldsymbol{b}+\boldsymbol{c}|^{2}=s^{2}+\left(t+\frac{s-1}{t}\right)^{2} \geqslant s^{2}+4(s-1)$ $\Rightarrow s^{2}+4(s-1) \leqslant 8 \Rightarrow(s-2)(s+6) \leqslant 0 \Rightarrow s \leqslant 2$, equality holds when $t= \pm 1$. Therefore, the maximum value of $\boldsymbol{a} \cdot \boldsymbol{c}$ is 2.