amparvardi wrote: Find all non-zero real numbers $x$ such that \[\min \left\{ 4, x+ \frac 4x \right\} \geq 8 \min \left\{ x,\frac 1x\right\} .\] The limits for casework are $0$ and the solutions of $x=\frac 1x$, so $x=-1,+1$ and of $4=x+\frac 4x$ and so $x=2$ 1) $x\le -1$ and the inequation becomes $x+\frac 4x\ge 8x$ $\iff$ $x^2\ge \frac 47$ and so $x\in(-\infty,-1]$ 2) $-1\le x<0$ and the inequation becomes $x+\frac 4x\ge \frac 8x$ $\iff$ $x^2\le 4$ and so $x\in[-1,0)$ 3) $0<x\le 1$ and the inequation becomes $4\ge 8x$ and so $x\in(0,\frac 12]$ 4) $1\le x\le 2$ and the inequation becomes $4\ge \frac 8x$ and so $x=2$ 5) $2\le x$ and the inequation becomes $x+\frac 4x\ge \frac 8x$ $\iff$ $x^2\ge 4$ and so $x\in[2,+\infty)$ Hence the answer : $\boxed{x\in(-\infty,0)\cup(0,\frac 12]\cup[2,+\infty)}$

1) If $x>0$, then $x+\frac{4}{x}\ge 4$, then $4\ge 8min\{x,\frac 1x\}$. It give $x\ge 2$ or $0<x\le 0.5$. 2) $x<0$, then $x+\frac{4}{x}\ge 8min\{x,\frac 1x\}$. 2.1. $-1\le x<0$ inequality hold. 2.2 $x<-1$. $x+\frac 4x\ge 8x\to x^2\ge \frac 47$, inequality hold $(-\infty,0), (0,0.5],[2,\infty).$

Remark $\boxed{\boxed{\min\{a,\ b\}=\frac{a+b}{2}-\frac{|a-b|}{2}}}$, $\min\left\{4,\ x+\frac{4}{x}\right\}\geq 8\min\left\{x,\ \frac{1}{x}\right\}\ (x\neq 0)$ $\Longleftrightarrow \frac{x+\frac{4}{x}+4}{2}-\frac{\left|x+\frac{4}{x}-4\right|}{2}\geq 8\left(\frac{x+\frac{1}{x}}{2}-\frac{\left|x-\frac{1}{x}\right|}{2}\right)$ $\Longleftrightarrow \frac{(x+2)^2}{x}-\left|\frac{(x-2)^2}{x}\right|\geq 8\left(\frac{x^2+1}{x}-\left|\frac{x^2-1}{x}\right|\right)$ $\Longleftrightarrow \frac{8|x^2-1|-(x-2)^2}{|x|}\geq \frac{8(x^2+1)-(x+2)^2}{x}$ $\Longleftrightarrow \frac{8|x^2-1|-(x-2)^2}{|x|}\geq \frac{7x^2-4x+4}{x}\ \cdots [*]$ Case 1: $x>0$ $[*]\Longleftrightarrow \frac{8|x^2-1|-(x-2)^2}{x}\geq \frac{7x^2-4x+4}{x}$ Since $x>0$, we have $|x^2-1|\geq x^2-x+1$, sketching two graphs of the curves $y=|x^2-1|,\ y=x^2-x+1$ gives the solution $0\leq x\leq \frac 12$ or $2\leq x$, since $x>0$, in this case, the solution is $0<x\leq \frac 12$ or $2\leq x\ \cdots [1]$. Case 2: $x<0$ $[*]\Longleftrightarrow \frac{(x-2)^2-8|x^2-1|}{-x}\geq \frac{7x^2-4x+4}{x}$ Since $x<0$, we have $8|x^2-1|-(x-2)^2\leq 7x^2-4x+4\Longleftrightarrow 8|x^2-1|\geq -6x^2$, which holds for all real numbers $x$. Thus in this case, the solution is $x<0\ \cdots [2]$. Therefore from $[1],\ [2]$, the desired answer is $\boxed{x<0\ or\ 0<x\leq \frac 12\ or\ 2\leq x}$