5. Find all values of $x$ for which the inequality $(a+2) x-(1+2 a) \sqrt[3]{x^{2}}-6 \sqrt[3]{x}+a^{2}+4 a-5>0$ holds for at least one $a \in[-2 ; 1]$.

#
# Solution:

$$
a^{2}+\left(x-2 \sqrt[3]{x^{2}}+4\right) a+2 x-\sqrt[3]{x^{2}}-6 \sqrt[3]{x}-5>0
$$

Find $x$, for which the inequality

$a^{2}+\left(x-2 \sqrt[3]{x^{2}}+4\right) a+2 x-\sqrt[3]{x^{2}}-6 \sqrt[3]{x}-5 \leq 0$ holds for all $a \in[-2 ; 1]$.

Let $f(a)=a^{2}+\left(x-2 \sqrt[3]{x^{2}}+4\right) a+2 x-\sqrt[3]{x^{2}}-6 \sqrt[3]{x}-5$. Then $\left\{\begin{array}{c}f(-2) \leq 0, \\ f(1) \leq 0,\end{array}\left\{\begin{array}{c}3 \sqrt[3]{x^{2}}-6 \sqrt[3]{x}-9 \leq 0, \\ 3 x-3 \sqrt[3]{x^{2}}-6 \sqrt[3]{x} \leq 0\end{array}\right.\right.$ $t=\sqrt[3]{x},\left\{\begin{array}{c}t^{2}-2 t-3 \leq 0, \\ t\left(t^{2} \leq-t-2\right) \leq 0,\end{array} \quad t \in\{-1\} \cup[0 ; 2], \quad x \in\{-1\} \cup[0 ; 8]\right.$.

For $x \in(-\infty ;-1) \cup(-1 ; 0) \cup(8 ;+\infty)$.

Answer: $(-\infty ;-1) \cup(-1 ; 0) \cup(8 ;+\infty)$