Áp dụng bđt bđt |a| + |b| + |c| \(\ge\) |a + b + c| ta có:
A = |2x - y| + |1 - 2x| + |y + 5| + |2x - 1| \(\ge\) |2x - y + 1 - 2x + y + 5 | + |2x - 1| = |6| + |2x - 1| = 6 + |2x - 1| \(\ge6\)
Dấu "=" xảy ra khi \(\left\{\begin{matrix}2x-y\ge0\\2x-1\le0\\y+5\ge0\\2x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}2x\ge y\ge-5\\2x=1\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}1\ge y\ge-5\\x=\frac{1}{2}\end{matrix}\right.\)
Ta có:
\(\left|2x-y\right|\ge0\forall x,y\\ \left|2x-1\right|\ge0\forall x\\ \Rightarrow2\left|2x-1\right|\ge0\forall x\\ \left|y+5\right|\ge0\forall y\)
\(\Rightarrow\left|2x-y\right|+2\left|2x-1\right|+\left|y+5\right|\ge0\forall x,y\)
\(hay:A\ge0\forall x,y\)
Vậy: \(Min_A=0\) tại \(x=2,5;y=-5\)