\(\sqrt{x^2+4x+4}+\sqrt{x^2-4x+4}\)
\(=\sqrt{\left(x+2\right)^2}+\sqrt{\left(x-2\right)^2}\)
\(=\left|x+2\right|+\left|x-2\right|\)
\(=\left|x+2\right|+\left|2-x\right|\ge\left|x+2+2-x\right|=\left|4\right|=4\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x+2\right)\left(2-x\right)\ge0\Leftrightarrow-2\le x\le2\)