Prove that, if $|\alpha| < 2\sqrt{2}$, then there is no value of $x$ for which $$x^2-\alpha|x| + 2 < 0\qquad\qquad(*)$$
Find the solution set of (*) for $\alpha=3$.
For $\alpha > 2\sqrt{2}$, then the sum of the lengths of the intervals in which $x$ satisfies (*) is denoted by $S$. Find $S$ in terns of $\alpha$ and deduce that $S < 2\alpha$.
The left side of (*) can be written as $$x^2-\alpha|x| + 2 = x^2 - 2\sqrt{2}|x| + 2 +(2\sqrt{2}-\alpha)|x| = (|x| -\sqrt{2})^2 + (2\sqrt{2}-\alpha)|x|$$
If $|\alpha| < 2\sqrt{2}$, then both terms will be non-negative. Hence (*) cannot hold.
Let $y=|x| \ge 0$. Then (*) is equivalent to $$y^2 - 3y + 2 < 0 \Leftrightarrow (y-1)(y-2) < 0 \Leftrightarrow 1 < y < 2 \Leftrightarrow 1 < |x| < 2$$
Hence, the solution is $\{1 < x < 2\}\cup\{-2 < x < -1\}$
Let $y_1 \le y_2$ be two roots to $y^2 - \alpha y + 2 =0$. Then, by Vieta theorem, we know $y_1 + y_2 = \alpha$ and $y_1y_2 = 2$. If $\alpha > 2\sqrt{2} > 0$, then both $y_1$ and $y_2$ are positive. This means that the solution to (*) will be $$\{-y_2, -y_1\}\cup\{y_1, y_2\}$$ The sum of their lengths is $$(-y_1 - (-y_2)) + (y_2-y_1)=2 (y_2-y_1)$$
By polynomial transformation and Vieta theorem, we have $$(y_2-y_1)^2 = (y_2+y_1)^2 - 4y_1y_2=\alpha^2 - 4\times 2 = \alpha^2 - 8$$
Hence $$2(y_2-y_1) = 2\sqrt{\alpha^2 - 8} < 2\sqrt{\alpha^2} = 2\alpha$$