Let $A = \{x \in \Bbb{R} : |2x -1| \leq 5\}$, let $B = \{x \in \Bbb{R} : |2x - 1| < 3\}$, let $D =\{x \in \Bbb{R} : x^2(7-x^2) \geq 7\}$ and let $E = \{x \in \Bbb{R} : 2x(x-2)(x+2) \geq 5 \}$. Prove by contradiction that $(A - B) \subseteq (D \cup E)$
I tried doing this in a sort of direct way but the problem got a little to complicated since it was hard to solve for $2x(x-2)(x+2) \geq 5$. I can't see any easy and elegant way to do it. I've also shown that $(A - B) = [-2, -1]\cup[2, 3]$. From there I tried to show how these two separate sets fit into D or E but I don't know if this is the proper way to solve this by contradiction.
Suppose to the contrary that the inclusion does not hold. Then there exists a point $x_0\in (A - B)\setminus (D \cup E)$. We already know that $A - B =[-2, -1]\cup[2, 3]$. Define functions $d$ and $e$ on $\Bbb R$ by putting $d(x)=x^2(7-x^2)-7$ and $e(x)=2x(x-2)(x+2)-5$ for each $x\in\Bbb R$. It is easy to check that $x\in D$ iff $d(x)\ge 0$ iff $\left|x^2-\tfrac 72\right|\le \tfrac{\sqrt{21}}{2}$. Since $\tfrac{\sqrt{21}}{2}\ge \tfrac 94$, we see that if $|x_0|\in \left[\tfrac{\sqrt{5}}2, \tfrac{\sqrt{23}}2 \right]$ then $x_0\in D$, a contradiction. If $x_0\ge \tfrac{\sqrt{23}}2$ then $e(x_0)\ge e\left(\tfrac{\sqrt{23}}2\right)= \tfrac{7\sqrt{23}}4-5>0$, so $x_0\in E$, a contradiction. If $-\tfrac{\sqrt{5}}2\le x_0\le -1$ then $e(x_0)\ge 2\left(4-\tfrac 54\right)-5>0$, so $x_0\in E$, a contradiction.