#Proving an inequality
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snaky_man
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So at some point you probably want to introduce a^2 + b^2 + c^2.
Yes, I am aware, the problem is, I haven't found a good way to do it
$|x| = \sqrt{x^2}$, then you can use the condition in terms like $(a+b+c)^2$, but it didn't get me anywhere
snaky_man
$a^2 + b^2 + c^2 = 1 \Rightarrow a^2,b^2,c^2 \in [0,1]$ this means that $a,b,c \in [-1,1]$ which allows some trigonometric substitution but this also doesn't seem to great
snaky_man
Okay, feels like you're barking up the wrong tree with all this.
I'm gonna tell you what I see.
What's |a + b + c|^2?
(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca)
There's your 1.
I know
that's what I meant here
using $|a+b+c| = \sqrt{(a+b+c)^2}$ is the easiest way to get to the $(a+b+c)^2$ term I guess
snaky_man
Disagree.
Also, that's not exactly why you have it.
At least, that's not the answer I was looking for.
wait, do you want me to explain why is there even a square root in the first place?
...I think so? I'm not quite sure what you mean, but do that and we'll see.
yes
What else is equal to |a + b + c| that has (a + b + c)^2 in it?
Right.
snaky_man
That's correct.
hmm
now it looks kind of like some form of cauchy-schwarz inequality
other than that I don't see how this helps
You know things like "Cauchy-Schwarz inequality" but you don't see how this helps?
yes?
is it really easy or what?
I mean, I just don't see how you can dismiss it out of hand. Like, it's not a magic bullet, but it's a start of a chain of logic.
okay, I will think about it
I still have no clue on how to proceed, can you give me a hint?
bruh got left on read
...start with |a + b + c| + |a + b - c| + |a + c - b| + |b + c - a|. Multiply by |a + b + c|. Divide by |a + b + c|.
To be honest, I don't know, I give up
Sad
Don't give up soldier
The fight starts once you feel like giving up
@steady root

...
I spent so much time on it and got nothing
I'm tired
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