ĐK: $x \geqslant - 1$
$\begin{array}
PT \Leftrightarrow {(2x)^3} + 2x < (x + 1 + 1)\sqrt {x + 1} \\
\,\,\,\,\,\,\, \Leftrightarrow {(2x)^3} + 2x < {(\sqrt {x + 1} )^3} + \sqrt {x + 1} \\
\,\,\,\,\,\,\, \Leftrightarrow f(2x) < f(\sqrt {x + 1} ),\,\,\,f(t) = {t^3} + t \\
\,\,\,\,\,\,\, \Leftrightarrow 2x < \sqrt {x + 1} \\
\,\,\,\,\,\,\, \Leftrightarrow \left[ \begin{array}
x < 0 \\
\left\{ \begin{array}
x \geqslant 0 \\
4{x^2} < x + 1 \\
\end{array} \right. \\
\end{array} \right. \Leftrightarrow \left[ \begin{array}
x < 0 \\
\left\{ \begin{array}
x \geqslant 0 \\
0 < x < \frac{{1 + \sqrt {17} }}{8} \\
\end{array} \right. \\
\end{array} \right. \\
\end{array} $
Vậy bất phương trình có nghiệm $ - 1 \leqslant x < \frac{{1 + \sqrt {17} }}{8}$