Absolute value problem

[tender stone]

Can someone help ≡(▔﹏▔)≡ ?
Let $a, b$ et $c$ be real numbers such that:
$$
|a-b| \geq|c|,|b-c| \geq|a| \text { et }|c-a| \geq|b| \text {. }
$$
Show that one of the three numbers a, b or $\mathrm{c}$ is the sum of the other two

[fierce island]

Ame-kun

[tender stone]

@tender stone

Hello

sorry for the ping

can you help?

Nah it's fine

Let me see

Give me a second

ok , I'm really thankful

Although, you have to ping the role associated with this topic, like <@&727457814523674674>, if your post wasn't answered in the first 15 minutes

oh oops

mb

ok

<@&727457814523674674>

Ok I have no idea how to solve this problem

You're on your own buddy

[tender stone]

damn

thanks

@dense inlet

Oh wait, you know the triangle inequality, right?

[dense inlet]

[tender stone]

|a| + |b| >= |a+b|, for any a,b real numbers

[dense inlet]

it is symmetry. We can WLOG |a|>=|b|>=|c|

[tender stone]

yea ofc I supposed that

[dense inlet]

Choose the one that is the sharpest $|b-c|\geq |a|$

[fierce island]

k12byda5h

[dense inlet]

wait lemme think sth

[tender stone]

I'm sorry for disturbing you on your gaming session

[dense inlet]

lol

[dense inlet]

$|b-c|\geq |a| \geq |b|$. Therefore, $sign(b) = - sign(c)$

[fierce island]

k12byda5h

[dense inlet]

(assume that non of a,b,c are 0 cuz it is minor case. So sign(a) is -1 or 1)

[tender stone]

all right

[dense inlet]

I stuck

[tender stone]

damn

my dad promised me some ice cream if I solve it

lol

[dense inlet]

wait wait. gimme a minute

[tender stone]

yea but if you find the answer don't tell me

I'm also thinking about a method to solve this problem

[dense inlet]

ok

I think I get it

[dense inlet]

what do you want me to tell

[tender stone]

an idea maybe?

[dense inlet]

square

[tender stone]

idk

[dense inlet]

$|x|\leq |y| \iff x^2 \leq y^2$

[fierce island]

k12byda5h

[tender stone]

ineffective

[dense inlet]

?

you can remove absolute. I think it is great

[tender stone]

okay, if I solve it I'll send you my answer

[dense inlet]

😉

[tender stone]

and I'll give u some ice cream too

[dense inlet]

how is it?

[tender stone]

I solved it

I'll send you the sol tomorrow

cuz it's night time now

[tender stone]

yesterday, but my mom took my laptop

good night

[tender stone]

@dense inlet

,w rotate

oh it doesn't rotate

[dense inlet]

My sol:\
square: $a^2+b^2-2ab-c^2 \geq 0$

[fierce island]

k12byda5h

[dense inlet]

equivalent to $(a-b+c)(-a+b+c)\leq 0$

[fierce island]

k12byda5h

[dense inlet]

$(a-b+c)(-a+b+c)\leq 0$\
$(a-b+c)(a+b-c)\leq 0$\
$(a+b-c)(-a+b+c)\leq 0$

[fierce island]

k12byda5h

[dense inlet]

obviously, one of them is 0, so we done

[tender stone]

@dense inlet

what do you think of mine?

[dense inlet]

ahhh

Any solutions that work are fine

[tender stone]

I just came back

from school

[dense inlet]

Your solution is what I thought at first.

It works too, but it might be exhaustin

g

[tender stone]

do I need to improve something?

[tender stone]

yours is more elegant

thx and sorry again for disturbing

[dense inlet]

On the first page, if all of them are positive, the inequality would be a<=c-b?

[tender stone]

no it's impossible for them to be all positive

cuz a is negative

and by symmetry they can't be all negative

[dense inlet]

yeah, just need to mention when writing full sol

[tender stone]

ah ok

[dense inlet]

but I'm too lazy to read lol

[tender stone]

sorry

[tender stone]

wth

lol

[tender stone]

ik my handwriting sucks

[dense inlet]

yk, I'm just lazy

[tender stone]

laziness is a personal trait for maths people