Có \(2x-2\sqrt{3x+1}-1\)
\(=\left(2x+\frac{2}{3}\right)-2\sqrt{\left(2x+\frac{2}{3}\right).\frac{3}{2}}+\frac{3}{2}-\frac{19}{6}\)
\(=\left(\sqrt{2x+\frac{2}{3}}-\sqrt{\frac{3}{2}}\right)^2-\frac{19}{6}\ge-\frac{19}{6}\forall x\ge-\frac{1}{3}\)
Dấu " =" xảy ra\(\Leftrightarrow\hept{\begin{cases}\sqrt{2x+\frac{2}{3}}=\sqrt{\frac{3}{2}}\\x\ge-\frac{1}{3}\end{cases}}\Leftrightarrow x=\frac{5}{12}\)
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