Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) cho \(VT\) ta có:
\(VT=\left|x+3\right|+\left|x-1\right|=\left|x+3\right|+\left|1-x\right|\)
\(\ge\left|x+3+1-x\right|=4\left(1\right)\)
Áp dụng tiếp BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) cho mẫu của \(VP\) ta có:
\(\left|y-2\right|+\left|y+2\right|=\left|2-y\right|+\left|y+2\right|\)
\(\ge\left|2-y+y+2\right|=4\)\(\Rightarrow\dfrac{1}{\left|y-2\right|+\left|y+2\right|}\le\dfrac{1}{4}\)
\(\Rightarrow VP=\dfrac{16}{\left|y-2\right|+\left|y+2\right|}\le\dfrac{16}{4}=4\left(2\right)\)
Từ \((1);(2)\) ta có: \(VT\ge4\ge VP\)
Đẳng thức xảy ra khi và chỉ khi \(VT=VP=4\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x+3\right|+\left|x-1\right|=4\\\dfrac{16}{\left|y-2\right|+\left|y+2\right|}=4\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\pm1\\x=-3\\x=-2\\x=0\end{matrix}\right.\\\left[{}\begin{matrix}y=\pm2\\y=\pm1\\y=0\end{matrix}\right.\end{matrix}\right.\)
