\(A=\sqrt{x^2+2x+1}+\sqrt{x^2-2x+1}=\sqrt{\left(x+1\right)^2}+\sqrt{\left(x-1\right)^2}=\sqrt{\left(x+1\right)^2}+\sqrt{\left(1-x^2\right)}\)
\(=\left|x+1\right|+\left|1-x\right|\)
Áp dụng bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(\left|x+1\right|+\left|1-x\right|\ge\left|x+1+1-x\right|=2\)
Dấu ''='' xảy ra khi \(x\le1\)