\(2=\left|a-1\right|+\left|b-1\right|\ge\left|a-1+b-1\right|=\left|a+b-2\right|\)
Với \(a+b-2\ge0\Leftrightarrow a+b\ge2\) thì:
\(a+b-2\le2\Leftrightarrow a+b\le4\Rightarrow\left|a+b-1\right|\le\left|4-1\right|=3\)
Dấu "=" xảy ra \(\left\{{}\begin{matrix}2\le a+b\le4\\a+b=4\\\left(a-1\right)\left(b-1\right)\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a\ge1,b\ge1\\a+b=4\end{matrix}\right.\)
Với \(a+b-2\le0\Leftrightarrow a+b\le2\) thì:
\(-a-b+2\le2\Leftrightarrow a+b\ge0\Rightarrow\left|a+b-1\right|\ge\left|0-1\right|=1\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a+b\le2\\a+b=0\\\left(a-1\right)\left(b-1\right)\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a\le1,b\le1\\a+b=0\end{matrix}\right.\)
Vậy GTNN của \(\left|a+b-1\right|\) là 1 khi \(\left\{{}\begin{matrix}a+b=0\\a\le1,b\le1\end{matrix}\right.\)
GTLN của $\left|a+b-1\right|$ là 3 khi $\hept{\begin{matrix}a\ge 1,b\ge 1\\a+b=4\end{matrix}}$
