5. Solve the inequality $\sqrt{(x+2)|x+1|+|x|} \geq x+2$.

#
# Solution:

$$
\left[\begin{array}{l}
\left\{\begin{array}{l}
x+2<0, \\
(x+2)(-x-1)-x \geq 0 ;
\end{array}\right. \\
\left\{\begin{array} { l } 
{ x + 2 \geq 0 , } \\
{ ( x + 2 ) | x + 1 | + | x | \geq ( x + 2 ) ^ { 2 } . }
\end{array} \Leftrightarrow \left[\begin{array}{l}
\left\{\begin{array}{l}
x<-2, \\
x^{2}+4 x+2 \leq 0 ;
\end{array}\right. \\
\left\{\begin{array}{l}
x \in[-2,-1]
\end{array}\right. \\
-x^{2}-3 x-2-x \geq x^{2}+4 x+4 ;
\end{array}\right.\right. \\
\left\{\begin{array}{l}
x \in(-1,0] \\
x^{2}+3 x+2-x \geq x^{2}+4 x+4 ;
\end{array}\right. \\
\left\{\begin{array}{l}
x \in(0,+\infty), \\
x^{2}+3 x+2+x \geq x^{2}+4 x+4
\end{array}\right.
\end{array} \Leftrightarrow\right.
$$

$\left[\begin{array}{l}\left\{\begin{array}{l}x<-2, \\ x \in[-2-\sqrt{2} ;-2+\sqrt{2}]\end{array}\right. \\ \left\{\begin{array}{l}x \in[-2,-1] \\ x^{2}+4 x+3 \leq 0 ; \\ x \in(-1,0], \\ 2 x+2 \leq 0 ;\end{array}\right. \\ \left\{\begin{array}{l}x \in(0,+\infty), \\ 2 \geq 4 .\end{array}\right.\end{array} \Leftrightarrow\left[\begin{array}{l}x \in[-2-\sqrt{2} ;-2) ; \\ x \in[-2,-1] ; \\ \text { no solutions; } \\ \text { no solutions. }\end{array} \Leftrightarrow x \in[-2-\sqrt{2} ;-1]\right.\right.$.

Answer: $x \in[-2-\sqrt{2} ;-1]$.