3:34 AM
@MrPie nice!

4 hours later…
7:05 AM
GarethMcCaughan ~ Catch a gamer hung @MrPie
Deusovi ~ devious -_- duh
El-Guest ~ eel guts ~ glue set
GarethMcCaughan ~ Catch a gamer hung
Mathphile ~ Phi Hamlet ~ I Help Math
Rand al'Thor ~ Ha! Rant lord!
Rubio ~ birou credits @Brandon_J
TheSimpliFire ~ Free Phi Limits
Omega Krypton ~ Gem Atop N. York

2 hours later…
9:06 AM
TheSimpliFire ~ This prime life.

indeed

GarethMcCaughan ~ Mr can cheat?! Uh, gag!

I think KevinL went for a long holiday?

@TheSimpliFire yeah, actually, where has he been lately?
Forgot the password to his account? ;)
(I hope not)
TheSimpliFire ~ Life's prime hit!
@TheSimpliFire

@MrPie was apparently seen 2 days ago

9:15 AM
@TheSimpliFire really? How do you check?

Kevin L, Singapore
2.3k 6 38
pretty simple

OOOOHHH
I had no idea you could do that!
BRUH
Didn't know it was just there

hehe, I do that all the time to see who's active and who's not

Well apparently you were last seen 12 minutes ago on Math SE

Takes time to update, though it should be seen 0 seconds ago as I'm still working on this!
(it is definitely one of the best problems on MSE)

9:31 AM
Lovely human being ~ gave in, only humble.
@TheSimpliFire I think I know a similar problem.
If $a^2+b^2+c^2+d^2\leqslant 2$ then prove that the sum of two values from $a,b,c,d$ is less than or equal to $\sqrt{2}$ (if I remember correctly)
I wonder if it can be generalised for $N$ where the sum of squares if $\leqslant N$ and we also have $\sqrt{N}$ instead of $2$ and $\sqrt{2}$
Perhaps we could apply this to what you're working on? idk

@MrPie so you want $a^2+b^2+c^2+d^2\le2\implies a+b\le\sqrt2$ wlog?

@TheSimpliFire I think so. I read this last year with my friend Harry and we couldn't prove it to be true
But I don't know if I remember it correctly. I think that's the one, though.

9:47 AM
$4(a^2+b^2+c^2+d^2)\ge(a+b+c+d)^2$ from C-S
so $(a+b+c+d)^2\le8$
@MrPie Are all of a,b,c,d positive?

@TheSimpliFire yes
I assume so

Then $a+b+c+d\le2\sqrt2$

@TheSimpliFire Well, $a\neq b\neq c\neq d$

So the inequality is strict

How come?

9:53 AM
Because equality at C-S holds when a/1 = b/1 = c/1 = d/1

...
...I have lost my mathematical touch, I reckon.

In mathematics, the Cauchy–Schwarz inequality, also known as the Cauchy–Bunyakovsky–Schwarz inequality, is a useful inequality encountered in many different settings, such as linear algebra, analysis, probability theory, vector algebra and other areas. It is considered to be one of the most important inequalities in all of mathematics.The inequality for sums was published by Augustin-Louis Cauchy (1821), while the corresponding inequality for integrals was first proved by Viktor Bunyakovsky (1859). The modern proof of the integral inequality was given by Hermann Amandus Schwarz (1888). ==...

Yeah, I know this ;)

Replace one of the terms by 1^2 + 1^2 + 1^2 + 1^2 and the rest follows

After this big anagram I'm working on, I'm gonna do some math

9:57 AM
We have $a+b\le\sqrt2,b+c\le\sqrt2,c+d\le\sqrt2,a+d\le\sqrt2$

Get back to my math methods book and just finish it so I CAN BLOODY MOVE ON IN SCHOOL
@TheSimpliFire you did it! :D
Can we generalise this for $N$?

@MrPie So you mean $a^n+b^n+c^n+d^n$?
Why four (a,b,c,d)?

@TheSimpliFire no I mean $a^2+b^2+c^2+d^2\leqslant N$ and the sum of two terms alone is $\leqslant \sqrt{N}$

@MrPie That's simple, just replace the '2' in the argument by 'N'

Exactly
Just reconfirming

10:02 AM
You have $a+b+c+d\le\sqrt{4N}$
But the other version of $a^n$ etc may be more difficult
We can use Holder's inequality for that actually
$$|a+b+c+d|\le(a^n+b^n+c^n+d^n)^{1/n}\cdot(1^{1-1/n}+1^{1-1/n}+1^{1-1/n}+1^{1-1/n})^{\frac1{1-1/n}}$$
to get $(a+b+c+d)^n\le 4^{n^2/(n-1)}(a^n+b^n+c^n+d^n)$
^^ should be $n-1$ instead of $n^2/(n-1)$
Essentially if $a^n+b^n+c^n+d^n\le n$ then $a+b+c+d\le4\sqrt[n]{\frac n4}$

10:20 AM
or $a+b+c+d\leq \sqrt[n]{n4^{n-1}}$

2 hours later…
12:03 PM
Deusovi ~ devious -_- duh
El-Guest ~ eel guts ~ glue set
GarethMcCaughan ~ Catch a gamer hung
Mathphile ~ Phi Hamlet ~ I Help Math
Rand al'Thor ~ Ha! Rant lord!
Rubio ~ birou credits @Brandon_J
TheSimpliFire ~ Free Phi Limits ~ This prime life [@MrPie] ~ Life's prime hit! [@MrPie]
Omega Krypton ~ Gem Atop N. York

"Catch a gamer hung" is much better than mine ;)
Guess there aren't any better ones

3 hours later…
3:08 PM
Rubio ~ I rob u.

3:28 PM
nice @Rubio
Deusovi ~ devious -_- duh
El-Guest ~ eel guts ~ glue set
GarethMcCaughan ~ Catch a gamer hung
Mathphile ~ Phi Hamlet ~ I Help Math
Rand al'Thor ~ Ha! Rant lord!
Rubio ~ birou credits @Brandon_J ~ I rob u credits @Rubio
TheSimpliFire ~ Free Phi Limits ~ This prime life [@MrPie] ~ Life's prime hit! [@MrPie]
Omega Krypton ~ Gem Atop N. York

:)