Note that we have two identities: \[ \left(1\right) \text{ } f\left(x\right)=\left(x^2-x+2\right)\left(ax+1-3a\right)+\left(a+1\right) \] \[ \left(2\right) \text{ } g\left(x\right)=\left(x^2-x+2\right)\left(\left(1-a\right)x-a\right)-\left(a+1\right) \] Suppose that there is a real $ z $ so that $ |f\left(z\right)|, |g\left(z\right)|<a+1 $ then we must have $ \left(z^2-z+2\right)\left(az+1-3a\right)<0 \text{ and } \left(z^2-z+2\right)\left(\left(1-a\right)z-a\right)>0 $ Because $ z^2-z+2=\left(z-\frac{1}{2}\right)^2+\frac{7}{4}>0 $, $ az+1-3a<0 \text{ and } \left(1-a\right)z-a>0 \Leftrightarrow \frac{a}{1-a}<z<\frac{3a-1}{a} $ However, it's impossible that $ \frac{a}{1-a}<\frac{3a-1}{a} \text{for some } a \in \left(0,1\right) $ since it's equivalent to $ \left(2a-1\right)^2<0 $, which is absurd! In conclusion, for any real $ x $, there is at least one of $ |f\left(x\right)| $ and $ |g\left(x\right)| $ isn't less than $ a+1 $