\(\left|x-1\right|+\left|x-2\right|+\left|y-3\right|+\left|x-4\right|=3\)
\(pt\Leftrightarrow\left|x-1\right|+\left|x-2\right|+\left|y-3\right|+\left|4-x\right|=3\)
Ta thấy: \(\left\{{}\begin{matrix}\left|x-2\right|\ge0\forall x\\\left|y-3\right|\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left|x-2\right|+\left|y-3\right|\ge0\forall x,y\left(1\right)\)
Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(\left|x-1\right|+\left|4-x\right|\ge\left|x-1+4-x\right|=3\left(2\right)\)
Cộng theo vế của \((1);(2)\) ta có:
\(VT=\left|x-1\right|+\left|x-2\right|+\left|y-3\right|+\left|x-4\right|\ge3=VP\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x-1\ge0\\x-2=0\\x-4\le0\\y-3=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge1\\x=2\\x\le4\\y=3\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
Vậy \((x;y)=(2;3)\) thỏa mãn
