Ta có: \(A=\sqrt{x^2-2x+1}+\sqrt{\left(x-4\right)^2}+\sqrt{\left(x-6\right)^2}\)
\(\Leftrightarrow A=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-4\right)^2}+\sqrt{\left(x-6\right)^2}\)
\(\Leftrightarrow A=\left|x-1\right|+\left|x-4\right|+\left|x-6\right|\)
Vì \(\left|a\right|=\left|-a\right|\) \(\Rightarrow\)\(\left|x-6\right|=\left|6-x\right|\)
Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)ta có:
\(\left|x-1\right|+\left|6-x\right|\ge\left|x-1+6-x\right|=5\)
\(\Rightarrow\)\(A\ge\left|x-4\right|+5\)
Vì \(\left|x-4\right|\ge0\forall x\)\(\Rightarrow\)\(\left|x-4\right|+5\ge5\forall x\)
\(\Rightarrow\)\(A\ge5\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-1\right)\left(6-x\right)>0\\x-4=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}1< x< 6\\x=4\end{cases}}\)
\(\Rightarrow x=4\)
Vậy \(A_{min}=5\)\(\Leftrightarrow\)\(x=4\)
