The minimum value of $1+8{\sin }^2{x}^2{\cos }^2{x}^2$  is

$\begin{array}{l}1+8{\sin }^2{x}^2{\cos }^2{x}^2\\ 1+8{\sin }^2{x}^2{\cos }^2{x}^2\\ =1+2.4{\sin }^2{x}^2.{\cos }^2{x}^2\\ =1+2.{\sin }^{2}2x^2\\ =1+[1-\cos \left(4x^2\right)]\\ =2-\cos \left(4x^2\right)\end{array}$

Minimum value$=c-\sqrt{a^2+b^2}$

$\begin{array}{l}=2-\sqrt{{(-1)}^2+{\left(0\right)}^2}\\ =2-1\\ =1\end{array}$