\[f^{'}(x) \leq g^{'}(x):\]
\[f(x) = x^{3} + x^{2} + x\sqrt{3};\]
\[g(x) = x\sqrt{3} + 1.\]
\[1)\ f^{'}(x) = 3x^{2} + 2x + \sqrt{3};\]
\[g^{'}(x) = \sqrt{3} + 0 = \sqrt{3}.\]
\[2)\ 3x^{2} + 2x + \sqrt{3} \leq \sqrt{3}\]
\[3x^{2} + 2x \leq 0\]
\[(3x + 2)x \leq 0\]
\[- \frac{2}{3} \leq x \leq 0.\]
\[Ответ:\ \ x \in \left\lbrack - \frac{2}{3};\ 0 \right\rbrack.\]