Ta có:
\(\left|2x+3\right|+\left|2x-1\right|=\left|2x+3\right|+\left|1-2x\right|\ge\left|2x+3+1-2x\right|=4\)(1)
\(\dfrac{8}{2\left(y-5\right)+2}\le\dfrac{8}{2}=4\) (2)
Mà: \(\left|2x+3\right|+\left|2x-1\right|=\dfrac{8}{2\left(y-5\right)^2+2}\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}-\dfrac{3}{2}\le x\le\dfrac{1}{2}\\y=5\end{matrix}\right.\)