Absolute value of addition of positive real numbers great than that of subtraction?

Question

$$∀a,b ∈ R+, |a + b| > |a - b|$$

I'm wondering if this is true? I'm not sure exactly how I could check or prove it to myself with the absolute value there. I thought I might be able to do something by squaring. Not sure if that works with absolute values.

Answer 1

It is true. You can prove it by cases. For instance, if $a > b$, then $a - b > 0$ and so $|a-b| = a- b$ and $|a+b| = a + b$. Similarly, you can check the other cases.

Answer 2

Since the absolute value of a number cannot be negative, the inequality is preserved if we square both sides.

\begin{align*} |a + b| & > |a - b|\\ |a + b|^2 & > |a - b|^2\\ (a + b)^2 & > (a - b)^2\\ a^2 + 2ab + b^2 & > a^2 - 2ab + b^2\\ 4ab & > 0 \end{align*}

The final inequality holds since $a, b \in \mathbb{R}^+$. Since the steps are reversible, the original inequality holds.