I have to solve different types of inequalites of this type:

$$|y^2-y-2|\geq 4 + |y^2+y-2|+|y+4|+|y|$$

I know the standard method for solving these inequalities, by finding the all zeros of the expressions and then analysing the sign of the expressions inside the absolute values for all the regions resulting from the zeros. But is there a smarter way to solve similar inequalities?

EDIT: The method can also use calculators but no advanced calculators (with plotting abilities or something like that.

  • $\begingroup$ The standard way is pretty much the cleverest way, too, I think. By the way, I don't think you mean "sign of the absolute values..." The sign of the absolute value is always positive or zero. :) $\endgroup$ – Thomas Andrews Jul 16 '16 at 13:42
  • $\begingroup$ I edited my op. I also think that there is no way around it, but maybe someone will share a neat trick to do this stuff faster :D. $\endgroup$ – MrYouMath Jul 16 '16 at 13:46
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    $\begingroup$ It happens that $$y^2-y-2=4 +(y^2+y-2)-(y+4)-y$$ hence, by the triangular inequality, the only way the desired inequality can hold is when it is an equality and every term involved has the same sign. Since one of these is $4$, which is positive, the inequality is equivalent to $$y^2-y-2>0\land y^2+y-2>0\land y+4<0\land y<0$$ Solving these is not difficult. $\endgroup$ – Did Jul 16 '16 at 14:48
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    $\begingroup$ *Typo: Replace strict inequalities in my comment by inequalities in the wide sense. $\endgroup$ – Did Jul 16 '16 at 15:02
  • $\begingroup$ @Did: wow, that was clever. Really liked this one :D $\endgroup$ – MrYouMath Jul 16 '16 at 15:04

When there is special structure, you can be clever. In your example, just saying $|y^2-y-2|\gt 4+ |y^2+y-2|$ forces $y$ to be greater than $2$ in absolute value. That means $y$ must also be negative to make the left side larger. That allows you to remove the absolute value bars on $y$ at the end, and to know that the long expressions are positive, leaving $y^2-y-2\geq 4 + y^2+y-2+|y+4|-y$ and $y \lt -2$. Now you can transform this to $-y \ge 4+|y+4|$ and $y+4$ must be negative, so we have $-y \ge 4$ or $y \le -4$.


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