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

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

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 ?

• Forget the $f$ and $g$; just think of it as $|a+b|=|a|+|b|\implies ab\ge0$. – Lord Shark the Unknown Mar 24 at 8:57
• ohh thanx. i think thts something i got into trouble thinking abt increasing and decreasing functions and all. So it'd be $$|f(x)+g(x)|^2=(|f(x)|+|g(x)|)^2\implies f^2(x)+g^2(x)+2f(x)g(x)=f^2(x)+g^2(x)+2|f(x)g(x)|\implies f(x)g(x)=|f(x)g(x)|\implies f(x)g(x)\geq 0$$,right – ss1729 Mar 24 at 9:08
• @ss1729 you can also take it as a proof by cases since there are only 4 as a describe in my answer. That’s just an alternate approach. – rb612 Mar 24 at 9:11
• @ss1729 That's good. – egreg Mar 24 at 10:39

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 $$b=g(x)$$.

• About a geometric interpretation, you can assume $a>0$ (the case $a=0$ is trivial and if $a<0$ just change $a$ into $-a$ and $b$ into $-b$). What happens if $b<0$? – egreg Mar 24 at 11:09

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.