$\quad 8-2x-x^2\ge0\Longrightarrow x\in[-4,2]$. $\quad\textbf{Case 1}: x\in[-4,0)$. The inequality becomes: $-5x\le x(3x+2-2\sqrt{8-2x-x^2})\Longleftrightarrow 3x+2-2\sqrt{8-2x-x^2}\le -5\Longleftrightarrow$ $\Longleftrightarrow 2\sqrt{8-2x-x^2}\ge 3x+7$. $\quad$ $\textbf{Case 1.1}$: $3x+7\le 0\Longrightarrow x\in\left[-4, -\dfrac{7}{3}\right]$. $\quad$ $\textbf{Case 1.2}$: $3x+7\ge 0$. Squaring, results: $4(8-2x-x^2)\ge 9x^2+42x+49\Longleftrightarrow 13x^2+50x+17\le0$, with the solution $x\in\left[\dfrac{-25-2\sqrt{101}}{13}, \dfrac{-25+2\sqrt{101}}{13}\right]$. $\quad$ Results for the case $1$ the solution: $x\in\left[-4, \dfrac{-25+2\sqrt{101}}{13}\right]$. $\quad\textbf{Case 2}: x=0$. Results $0\le0$, true. Hence, for the case $2$, the solution is $x=0$. $\quad\textbf{Case 3}: x\in(0,2]$. The inequality becomes: $5x\le x(3x+2-2\sqrt{8-2x-x^2})\Longleftrightarrow 3x+2-2\sqrt{8-2x-x^2}\ge 5\Longleftrightarrow$ $\Longleftrightarrow 2\sqrt{8-2x-x^2}\le 3x-3$. $\quad$ $\textbf{Case 3.1}$: $x\in(0,1)\Longrightarrow 2\sqrt{8-2x-x^2}\le 3x-3<0$, no solution. $\quad$ $\textbf{Case 3.2}$: $x\in[1,2]$. Squaring, results: $4(8-2x-x^2)\ge 9x^2-18x+9\Longleftrightarrow 13x^2-10x-23\ge0$, with the solution $x\in\left[\dfrac{23}{13},2\right]$. $\quad\textbf{Final answer}$: $\quad x\in \left[-4, \dfrac{-25+2\sqrt{101}}{13}\right]\cup\{0\}\cup \left[\dfrac{23}{13},2\right]$.

gobathegreat wrote: Solve the inequation: $$5\mid x\mid \leq x(3x+2-2\sqrt{8-2x-x^2})$$ It's Lomonosov Olimpiad 2005, see here http://mathus.ru/olymp/lom2005m.pdf I solved this problem when I was at school