Absolute value inequalities with quadratics.

Question

I came across the following question in a book:

$4|x^2-1|+|x^2-4| \ge 6$

Now, the general method to solve inequalities is to break it into cases and then solve it. However, that method is quite lengthy and takes up time. Here, in our classes(coaching classes), we're expected to solve such question in less than 2 minutes. So, I was wondering if there are any dodges/tricks to solve such questions more easily. I just presented this question as an example, but I would like a method that can be employed in number of general cases. Any help will be appreciated.

BTW, the answer to this is: $x \in (- \infty, - \sqrt2] \cup[- \sqrt{\frac 2 5}, + \sqrt{\frac 2 5}] \cup [+ \sqrt2, \infty) $

Thanks!

Answer 1

Let $x^2=t$.

Thus, we need to solve $f(t)\geq6,$ where $$f(t)=4|t-1|+|t-4|.$$ We see that $f$ is a convex function (why?),

which says that we need to solve: $t\geq t_1$ or $t\leq t_2$, where $t_1\geq t_2$ and $t_1$ and $t_2$ are roots of the equation: $$f(t)=6,$$ which gives $t_1=2$, $t_2=\frac{4}{5}$ and we obtain your answer during $30$ seconds.

Answer 2

The eqn. is $$4|x^2-1|+|x^2-4|\ge 6 ~~~~~~(*)$$ Let $z=x^2$, then the eqn, is $$4|z-1|+|z-4| \ge 6$$ Three regions of $z$ hace to be explored.

$R_1: z \le 1:$ then $-4(z-1)+-(z-4) \ge 6 \implies z \le 2/5 \implies x^2 \le 2/5 \implies x \in [-\sqrt{2/5},\sqrt{2/5}]~~~~(1)$.

$R_2: 1< z \le 4: 4(z-1)+-(z-4) \ge 6 \implies 2 \le z \le 4 \implies 2 \le x^2 \le 4 \implies x \in [-2,\sqrt{2}] \cup [\sqrt{2}, 2].~~~~(2)$

$R_2: z >4: 4(z-1)+(z-4) \ge 6 \implies z >14/5 \implies z \ge 4 \implies x^2 >4 \implies x \in (-\infty, -2) \cup (2, \infty).~~~~(3)$

The final solution id the union of (1), (2), (3) as: $$x\in (-\infty, -\sqrt{2}] \cup [-\sqrt{2/5}, \sqrt{2/5}] \cup [\sqrt{2}, \infty)$$, See the Fig. below where Eq. (*) is writren as $f(x)=6$ and $y=f(x)$ and $y=6$ are plotted, notice three intervals where $f(x)\ge 6$.