\(\begin{array}{l}
\left| {\left| {{x^2} - 4x - 5} \right| + 2x + 9} \right| \le \left| {{x^2} - x + 5} \right|\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left| {{x^2} - 4x - 5 + 2x + 9} \right| \le \left| {{x^2} - x + 5} \right|\\
\left| { - {x^2} + 4x + 5 + 2x + 9} \right| \le \left| {{x^2} - x + 5} \right|
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left| {{x^2} - 2x + 4} \right| \le \left| {{x^2} - x + 5} \right|\\
\left| { - {x^2} + 6x + 14} \right| \le \left| {{x^2} - x + 5} \right|
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left( {2{x^2} - 3x + 9} \right).\left( { - x - 1} \right) \le 0\\
\left( {5x + 19} \right).\left( { - 2{x^2} + 7x + 9} \right) \le 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - 1\\
 - \frac{{19}}{5} \le x \le  - 1 \vee x \ge \frac{9}{2}
\end{array} \right. \Leftrightarrow x \ge \frac{9}{2}
\end{array}\)