\(D=\sqrt{\left(2x-1\right)^2+16}+\left|y^2+1\right|+2\)
Ta có:\(\left\{{}\begin{matrix}\sqrt{\left(2x-1\right)^2+16}\ge\sqrt{16}=4\\\left|y^2+1\right|\ge1\end{matrix}\right.\)
Nên:\(D\ge4+1+2=7\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=0\end{matrix}\right.\)