Prove and interpret $|f(x)+g(x)|=|f(x)|+|g(x)| \implies f(x)g(x)\geq0$

Question

$$ |f(x)+g(x)|=|f(x)|+|g(x)| \implies f(x)g(x)\geq0 $$

I don't have any clue of how to prove this ?

Can someone give any geometrical interpretation to it, as I really don't want to just bihart it ?

Answer 1

The functions are a red herring. You just have to prove that, for any real $a$ and $b$,

if $|a+b|=|a|+|b|$, then $ab\ge0$.

If $|a+b|=|a|+|b|$, then $|a+b|^2=(|a|+|b|)^2$, which translates into $$ a^2+2ab+b^2=a^2+2|ab|+b^2 $$ hence $ab=|ab|$, which is equivalent to $ab\ge0$.

Now set $a=f(x)$ and $b=g(x)$.

Answer 2

Hint: it might help to think about what cases the implication holds true. $f(x)$ and $g(x)$ are just real numbers (assuming that this is the function range), so what are the possibilities for $a$ and $b$ so that $ab \geq 0$? Think about the sign (positive/negative) of $a$ and $b$.

Now think about how this relates to absolute values. Think about what happens if you plug in a combination of positive or negative values to the absolute value equation. In what cases does it hold true?

Interpreting geometrically, think of $1$ as being a step to the right on the number line, and $-1$ as being a step to the left. What cases are the number of steps you take the same as the end distance you are from $0$? For example, 5 steps to the right and 2 steps to the left brings me to a net total of 3 steps to the right. But I took 7 steps total. So the absolute value equation doesn’t hold true.