Đáp án:
\(\min C = 2\Leftrightarrow x =0\)
Giải thích các bước giải:
\(\begin{array}{l}
C = \sqrt{x^2 + x + 1} + \sqrt{x^2 - x + 1}\\
\to C = \sqrt{\left(x + \dfrac12\right)^2 + \dfrac34} + \sqrt{\left(x - \dfrac12\right)^2 + \dfrac34}\\
\to C = \sqrt{\left(x + \dfrac12\right)^2 + \left(\dfrac{\sqrt3}{2}\right)^2} + \sqrt{\left(\dfrac12-x\right)^2 + \left(\dfrac{\sqrt3}{2}\right)^2}\\
\to C \geqslant \sqrt{\left( x + \dfrac12 + \dfrac12 - x\right)^2 + \left(\dfrac{\sqrt3}{2} + \dfrac{\sqrt3}{2}\right)^2}\quad (BDT\ Minkowski)\\
\to C \geqslant \sqrt{1 + 3}\\
\to C \geqslant 2\\
\text{Dấu = xảy ra}\ \Leftrightarrow \dfrac{x+\dfrac{1}{2}}{\dfrac{\sqrt3}{2}} = \dfrac{\dfrac{1}{2}-x}{\dfrac{\sqrt3}{2}}\Leftrightarrow x=0\\
\text{Vậy}\ \min C = 2 \Leftrightarrow x = 0
\end{array}\)
