Can I replace modulus inequalities with rooted square arguements?

Question

Suppose I want to show $|x-5|<|x+1|$. One way (and the way my lecturer shows) to do it is look at the negative and positive regions and solve the inequality. But with the definition $\sqrt{x}\geq0$, $\forall x\in \Bbb R$, I can say that $|x|=\sqrt{x^{2}}$. So could I, in general over $\Bbb R$, use this definition to substitute the modulus argument? Obviously in some cases this won't do anything because I will end up with the same +/- situation, but at least here the task is (at least for my brain) more algebraically simple, because: $$ |x-5|<|x+1| \iff\sqrt{(x-5)^{2}} < \sqrt{(x+1)^{2}}\iff(x-5)^{2}<(x+1)^{2}$$ With dif. squares:$$(x-5)^{2}-(x+1)^{2}<0$$ $$(2x-4)(-6)<0$$ $$x>2$$ To me this seems extremely obvious because I can use the "intuition" $$a<b, c<d\implies ac<bd \quad\forall a,b,c,d \in \Bbb R_{\geq 0 },$$ but I'm new to real analysis and everything seems to have weird caveats that make nothing intuitive true in general haha. Is there a proof for this or is the intuition something that just logically extends from Peano's axioms and natural number theorems?

Answer 1

The fact that $|x| = \sqrt{x^2}$ is proof enough for you to use the method you suggested here. As you already noted, sometimes this method is easier, but sometimes you will end up with something more complicated (say, if the original question had been $|x-5|<|2x+2|$, for example). In general, it's good to understand and be able to use multiple methods for solving algebraic equations.