Có
\(\left|x-2\right|+\left|x-4\right|=\left|x-2\right|+\left|4-x\right|\ge\left|x-2+4-x\right|=2\)
\(\left|x-3\right|\ge0\)
=> \(\left|x-2\right|+\left|x-4\right|+\left|x-3\right|\ge2\)
Dấu "=" xảy ra
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x-2>0\\4-x>0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=3\\x-2< 0\\4-x< 0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\x>2\\x< 4\end{matrix}\right.\\\left\{{}\begin{matrix}x=3\\x< 2\\x>4\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow x=3\)
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