I can AM-GM it for integral p, q

@bolbteppa Well that's not the full story, because the notation $(n_1,n_2)$ refers to a representation.

Not in general, you mean.

I keep forgetting that identity from my short term memory, but a non-calculus way of looking at it is that it's expressing the concavity of the logarithm

@Blue There's probably subtleties in general. You have to prove for rational $p, q$'s, which should be only slightly harder

Then by density of rationals extend that to $p, q \in \Bbb R$

@bolbteppa I think it's the Holder inequality

@Blue Yeah simplest is to prove it using Lagrange mutipliers

This is Young's inequality which you use to prove Holder's inequality right

Oh yes sorry

I forget names

Of course calculus lets you prove the log is concave quick enough

and the whole 1/p, 1/q business is magic from this perspective

the integral proof is another way of seeing the 1/p, 1/q stuff more naturally, and the picture is cool

proofwiki.org/wiki/Young%27s_Inequality_for_Products#Proof_2

I agree, but if you do it the other way the p's and 1/p's are trickery

@Blue Do this

I know another proof which is too crazy to try to remember the details of

@Blue Am Gm,

and then derive holders from it

@PrathyushPoduval Only works for integral $p, q$

@davidphysics yes everything I have read is indeed very hand-wavey talking about this, not clearly explaining how it is a representation, just stating it

But it can be made to work for rational $p, q$ like I said, and you'd prove the full thing from it

@BalarkaSen no it needdt

I think am gm extends for genreal t and 1-t

@PrathyushPoduval Sure, same as my suggestion

ah yeas

But good to know that works

This on Young looks interesting

math.upenn.edu/~brweber/Courses/2011/Math361/Notes/YMandH.pdf

@Blue Why are you differentiating wrt $p$ and $q$? That's not what you want to maximize...

You have fixed $p$ and $q$

You want to maximize $f(\alpha, \beta) = \alpha^p/p + \beta^q/q$

Just like $0 \leq (a - b)^2 = a^2 + b^2 - 2ab \to ab \leq \frac{1}{2}a^2 + \frac{1}{2}b^2$ you can write Young as a generalization $ab \leq \frac{p-1}{p} a^{p/(p-1)} + \frac{1}{p} b^p$

use jensens inequality

@Blue^

It's too arbitrary to add in these $(p-1)/p$ terms :(

do you want a calculus methos>

?

okay lets see

1) there are no constrains on your variables

@Blue You wouldn't get that

$1/p + 1/q = 1$ is not a constraint on $\alpha, \beta$

@PrathyushPoduval You have to set up one

Try $\alpha \beta = C$

yes

That's not a constraint.

It's a condition on $p$ and $q$.

A constraint is a condition on the variables when you're extremizing something

Eh, it's a constraint on the parameters defining the problem instead of on independent variables.

The condition "1/p+1/q=1".

I'm saying that the only real difference between 'condition' and 'constraint' here is that the former applies to the parameters of the problem and the latter to the independent variables of the problem.

From $ab = e^{\ln(ab)}$ you can reduce the problem of analyzing $ab$ to analyzing the convexity of $\ln(x)$ (or $e^x$ if you want), then use the above.

@BalarkaSen I have all your movies

Wee

I'll set up an access server today

I wonder if it'll take infinitely long to download on my end

Possibly

How big are these lmao

@JohnRennie Can you lend me a VM with some 10GB of storage so I can transfer some files to @BalarkaSen without having to leave my laptop on at all times?

@BalarkaSen Totalling some 10GB I think

Fucking hell

I suspect this will take days

If John can't help I'll set up my raspberry pi as a provider

That's fine

Assuming my internet doesn't die, lol

Same!!!

We just need to use rsync

Oh wait

You run malware right?

lol

i run windows

Sigh

Why dude

top 10 anime betrayals

Let's wait for @JohnRennie's advice, I don't know how to do anything reliable on Windows

Going to shower; brb

See ya

my download is like 20 mbps.

@Blue Not sure what happens

@davidphysics Yes, you are correct. But I also think this question is a sufficient duplicate for that - you could add your answer there.

@ACuriousMind Halp

@BernardoMeurer Please review my melon joke that the math chat is incapable of understanding

@BernardoMeurer ?

@ACuriousMind Ah, I don't remember what I need help with now

@ACuriousMind but please do help me with it

@BernardoMeurer what do you need

If you ask a similar variant of the same question multiple times on a site you will get multiple answers from different people who see it in the moment, if you only allow it to be asked once you will get maybe one answer and people wont answer it years later as it looks closed and answered, a policy closing these kinds of questions wrecks the site, and lets it's competitiors e.g. the math version of this, do better

person is talking about thermo and EM here, physics.stackexchange.com/q/372185/25851, linked post physics.stackexchange.com/q/70376/25851 is going on about lagrangians, makes no sense to think this answers it, ridiculous ridiculous policy

@0celo7 I already solved all those proofs but now I have a new one

@BernardoMeurer what is it

@0celo7 It involves geometry, I'm drawing it on a paper

ew

I sent it to you on iMessage

it's hideous

I don't even know what it is tbh

@BernardoMeurer I don't believe it for the example you drew

@0celo7 Me neither

there are 8 tiles, which is not div by 3

I got it from that picture I sent you yesterday

I'm missing a row of triagles I thinkg

maybe, I dunno

the picture cuts off

Ye

If I add another row it's 16 triangles ($4^2$) and if you take the edge of it's 15, which is div. by 3

ok

it's some horrible induction proof then

Is it not just showing that any $4^n - 1$ is divisible by 3?

maybe, but you need to show that it's equivalent to that

I guess I need to show that any three conjoined eq. triangles is a trapezoid ?

that's not true

Counter example?

oh, equilateral

yeah that's true

Ye

How do I show that even

draw all the combinations

Any way to do it analytically?

this isn't analysis

;-;

I don't want to draw

I'll say it's trivial