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

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 ?

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    $\begingroup$ Forget the $f$ and $g$; just think of it as $|a+b|=|a|+|b|\implies ab\ge0$. $\endgroup$ – Lord Shark the Unknown Mar 24 at 8:57
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    $\begingroup$ 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 $\endgroup$ – ss1729 Mar 24 at 9:08
  • $\begingroup$ @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. $\endgroup$ – rb612 Mar 24 at 9:11
  • $\begingroup$ @ss1729 That's good. $\endgroup$ – 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 $b=g(x)$.

  • $\begingroup$ 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 $b<0$? $\endgroup$ – 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.


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