Squaring without splitting into cases

In the two previous cases the splitting arose from the fact that the right-hand side could have any sign. If instead the right-hand side is, by construction, a quantity that is always $\ge 0$ — another radical, an absolute value, a sum of squares, a distance — then the sign study is free: there is no need to distinguish cases. It is enough to impose the existence conditions and square directly.

Property — When one can square without splitting into cases

Given an inequality of the form $\sqrt{A(x)}\ \lessgtr\ \sqrt{B(x)},\qquad \sqrt{A(x)}\ \lessgtr\ |C(x)|,\qquad |A(x)|\ \lessgtr\ |C(x)|,$ the right-hand side is always $\ge 0$ (within the existence conditions). Therefore the left-hand side too, being non-negative, has the same sign and one can square directly without distinguishing cases. The procedure reduces to three steps:

1. Write the existence conditions (on the radicals on the left- and right-hand sides).
2. Square both sides.
3. Intersect the solution with the existence conditions.

Proof

For the equivalence $\sqrt{A}\le |C| \iff A\le C^2$ within the existence conditions it suffices to observe that both sides are $\ge 0$: the function $t\mapsto t^2$ is strictly increasing on $[0,+\infty)$ and hence monotonic; applying it to each side preserves the inequality. For $|A|\le |C|$ the reasoning is identical, noting that $|A|^2=A^2$ and $|C|^2=C^2$: the absolute value disappears the moment one squares, and need not even be “studied”. $\blacksquare$

Links

Topics: Irrational inequalities
Concepts: Existence conditions · Irrational inequality · Absolute value
Methods: Squaring · Non-negative sides shortcut
Skills: Proving · Solving inequalities