OK's 屋企(home) - 2019-06-22

3:34 AM

@MrPie nice!

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

Mr Pie

9:06 AM

TheSimpliFire ~ This prime life.

TheSimpliFire

indeed

Mr Pie

GarethMcCaughan ~ Mr can cheat?! Uh, gag!

TheSimpliFire

I think KevinL went for a long holiday?

Mr Pie

@TheSimpliFire yeah, actually, where has he been lately?

Forgot the password to his account? ;)

(I hope not)

TheSimpliFire ~ Life's prime hit!

@TheSimpliFire

TheSimpliFire

@MrPie was apparently seen 2 days ago

Mr Pie

9:15 AM

@TheSimpliFire really? How do you check?

TheSimpliFire

Kevin L, Singapore

2.3k 6 38

pretty simple

Mr Pie

OOOOHHH

I had no idea you could do that!

BRUH

Didn't know it was just there

TheSimpliFire

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

Mr Pie

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

TheSimpliFire

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)

Mr Pie

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

TheSimpliFire

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

Mr Pie

@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.

TheSimpliFire

9:47 AM

Nvm, I misread the implication

$4(a^2+b^2+c^2+d^2)\ge(a+b+c+d)^2$ from C-S

so $(a+b+c+d)^2\le8$

@MrPie Are all of a,b,c,d positive?

Mr Pie

@TheSimpliFire yes

I assume so

TheSimpliFire

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

Mr Pie

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

TheSimpliFire

So the inequality is strict

Mr Pie

How come?

TheSimpliFire

9:53 AM

Because equality at C-S holds when a/1 = b/1 = c/1 = d/1

Mr Pie

...

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

TheSimpliFire

Cauchy–Schwarz inequality

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). ==...

Mr Pie

Yeah, I know this ;)

TheSimpliFire

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

Mr Pie

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

TheSimpliFire

9:57 AM

We have $a+b\le\sqrt2,b+c\le\sqrt2,c+d\le\sqrt2,a+d\le\sqrt2$

Mr Pie

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 $N$?

TheSimpliFire

@MrPie So you mean $a^n+b^n+c^n+d^n$?

Why four (a,b,c,d)?

Mr Pie

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

TheSimpliFire

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

Mr Pie

Exactly

Just reconfirming

TheSimpliFire

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 $a+b+c+d\le4\sqrt[n]{\frac n4}$

Mr Pie

10:20 AM

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

Omega Krypton

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

Mr Pie

"Catch a gamer hung" is much better than mine ;)

Guess there aren't any better ones

Rubio

3:08 PM

Rubio ~ I rob u.

Omega Krypton

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

Rubio

:)

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