How to simplify an expression with absolute and log functions?

Question

I'm confused with regard to simplifying this expression:

$$ |x| - |x-A| > \ln(\Gamma) $$

I was thinking of taking square on both ends, and that's basically where I got confused. Should I square the whole left side of the expression, i.e. $(|x| - |x-A|)^2$ or can I get away with just squaring them individually?

If I decide to square them individually, I get the following:

$$ 2Ax - A^2 > (\ln(\Gamma))^2 $$

Finally:

$$ x > (\ln(\Gamma)^2 + A^2)/2A $$

Am I doing this correctly?

Answer 1

When you have $$a<b$$ you can not always just square both sides. You must have that both sides are positive which does not seem to be your case.

But you might try the following: You have $$|x| - |x-A| \leq |A|$$ So $$ |A| \geq \ln (\Gamma).$$

I don't know if this is good enough for your purposes.

Answer 2

No, you can't square terms being added together individually. For instance, this is true: $$1+1=2,$$ but this is not: $$1^2+1^2=2^2.$$

Depending on what you want, you could make a number of simplifications. One would be to multiply by $|x|+|x-A|$, creating a difference of squares on the left to get: $$x^2-(x-A)^2 > \ln(\Gamma)(|x|+|x-A|).$$

If you want to get rid of every absolute value, then you'll have to square multiple times. For instance, assuming both sides are positive, you can square the whole expression to get $$x^2+(x-A)^2 -|x(x-A)|>\ln(\Gamma)^2$$ And then rearrange to get $$x^2+(x-A)^2-\ln(\Gamma)^2>|x(x-A)|$$ and then square again (since both sides are positive now, assuming their were initially) $$(x^2+(x-A)^2-\ln(\Gamma)^2)^2>x^2(x-A)^2$$ which has no absolute values.

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However, algebraic manipulations are not a particularly good way to attack this sort of problem. If we split into cases, then we can get rid of the absolute values entirely; for instance, if we assume $A$ is positive (it's easy to work out the necessary modifications if it negative) we split into cases:

• Case 1: $x\leq 0$. Then $x-A<0$ so $|x|=-x$ and $|x-A|=-x+A$. Thus the inequality is $-A > \ln(\Gamma)$ in this case.
• Case 2: $0\leq x \leq A$. Then $|x|=x$ and $|x-A|=-x+A$. The inequality is $2x - A > \ln(\Gamma)$ in this case.
• Case 3: $A \leq x$. Then $|x|=x$ and $|x-A|=x-A$. The inequality is $A>\ln(\Gamma)$ in this case.

We can quickly work out constraints on $x$ assuming $A$ is positive - in particular, if $-A>\ln(\Gamma)$, then all $x$ satisfy the inequality. If $A\leq \ln(\Gamma)$, then no $x$ satisfy the inequality. If $-A\leq \ln(\Gamma)<A$, then any $x$ greater than $\frac{\ln(\Gamma)+A}2$ satisfies the inequality.