Đặt \(A=\sqrt{x^2+2x+1}+\sqrt{x^2-4x+4}\)

\(A=\sqrt{\left(x+1\right)^2}+\sqrt{\left(x-2\right)^2}\)

\(A=\left|x+1\right|+\left|x-2\right|\)

\(A=\left|x+1\right|+\left|2-x\right|\)

Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)ta có :

\(A=\left|x+1\right|+\left|2-x\right|\ge\left|x+1+2-x\right|=\left|3\right|=3\)

Đẳng thức xảy ra khi ab ≥ 0

=> ( x + 1 )( 2 - x ) ≥ 0

Xét hai trường hợp :

1. \(\hept{\begin{cases}x+1\ge0\\2-x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge-1\\-x\ge-2\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge-1\\x\le2\end{cases}}\Leftrightarrow-1\le x\le2\)

2. \(\hept{\begin{cases}x+1\le0\\2-x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le-1\\-x\le-2\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le-1\\x\ge2\end{cases}}\)( loại )

=> MinA = 3 <=> \(-1\le x\le2\)

\(\sqrt{2}A=\sqrt{4x^2-4x+10}+\sqrt{4x^2-8x+8}\)

\(\sqrt{2}A=\sqrt{\left(2x-1\right)^2+3^2}+\sqrt{\left(2-2x\right)^2+2^2}\)

Áp dụng BĐT \(\sqrt{A^2+B^2}+\sqrt{C^2+D^2}\ge\sqrt{\left(A+C\right)^2+\left(B+D\right)^2}\)

=>\(\sqrt{2}A\ge\sqrt{\left(2x-1+2-2x\right)^2+\left(3+2\right)^2}=\sqrt{26}\)

=>\(A\ge\sqrt{13}\)

Dấu bằng xảy ra<=> \(\frac{2x-1}{3}=\frac{2x-2}{2}\)

<=>.........