National Math Olympiad

Let us check when $4\le x+\frac{4}{x}$. If $x>0$, then this inequality is equivalent to $4x\le x^2+4$ or $0\le (x-2{)}^2$. This is always true. If $x<0$, then we get $0\ge (x-2{)}^2$. This is never the case, so $4\ge x+\frac{4}{x}$. Hence, $$\min \left\{4,x+\frac{4}{x}\right\}=\{\begin{array}{ll}4,& \text{if }x>0,\\ x+\frac{4}{x},& \text{if }x<0.\end{array}$$ For $x>\frac{1}{x}$ and $x>0$ we get $x^2>1$, so $x>1$. For $x>\frac{1}{x}$ and $x<0$ we get $x^2<1$, so $-1<x<0$. Hence, $$\min \left\{x,\frac{1}{x}\right\}=\{\begin{array}{ll}\frac{1}{x},& \text{if }x>1\text{ or }-1<x<0,\\ x,& \text{if }x\le -1\text{ or }0<x\le 1.\end{array}$$ In the case of $x>1$ the given inequality is equivalent to $4\ge \frac{8}{x}$, which implies $x\ge 2$. The inequality holds for all numbers $x\in [2,\infty )$. If $0<x\le 1$, then we have $4\ge 8x$, so $\frac{1}{2}\ge x$. Thus, the inequality also holds for $x\in (0,\frac{1}{2}]$.

Let $-1<x<0$. We get $x+\frac{4}{x}\ge \frac{8}{x}$, or, equivalently, $x\ge \frac{4}{x}$. Multiplying by $x$ yields $x^2\le 4$, and this holds for $-1<x<0$. The inequality holds for all $x\in (-1,0)$.

Finally, let $x\le -1$. We get $x+\frac{4}{x}\ge 8x$ or $4\le 7x^2$. Since $x\le -1$, we have $x^2\ge 1$, so $7x^2\ge 7>4$. The inequality holds for all $x\in (-\infty ,-1]$ as well.

We have shown that the given inequality holds for all $x$ in $(-\infty ,0)\cup (0,\frac{1}{2}]\cup [2,\infty )$.