@0celo7 Yes

@BalarkaSen So I was thinking about this a bit. If $a,b$ are constants, such that $a,b>1$ or $a,b<1$, then there is no problem and Lagrange multipliers tell us that $ab\leq a^{p}/p+b^{q}/q$ and that greatest lower bound is attainable only when $a^{p}=b^{q}$. However, when $a>1,b<1$ or $a<1,b>1$ there is a problem since $a^{p}$ can never be equal to $b^{q}$ even after adjusting the positive $p,q$.

Thus, the Young's inequality is true even if $a,b$ are constants and $p,q$ are the variables given the constraint $p^{-1}+q^{-1}=1$.

According to the diagram, even when $a>1,b<1$: $ab$ will be a lower bound for sure. But it won't be the greatest lower bound from what I see.

Hello, everyone :-)

Hi!

@BalarkaSen That's what I was trying to do

As I mentioned yesterday

@BalarkaSen I'm considering $g(p,q)$ and trying to extremize it

I'm not proving Young's inequality, no

@BalarkaSen Well, if $a>1$ and $b<1$ then the minimum given by Holder's inequality is not attainable.

Or vice versa

Because the $a^{p}=b^{q}$ will never be true.

Yes, the inequality still holds. I was analyzing the various possibilities.

@JohnRennie Dang :-/

And it is equivalent to the restriction on AM-GM i.e. the extremum value is attained only when the terms are all equal

But sometimes all the terms cannot be equal in the AM-GM, and in those cases the minimum value is never attainable

Basically Young's is the Weighted AM-GM. Don't know if the exercise was worthwhile or not, but it made things clear for me.

@JohnRennie All done?

Cool :-)

Ah, well :-/

Right :-)

As per usual, then?

Yes, you've explained it before:

:-)

:-)

I miss home :-/

@JohnRennie Haha, right, right.

At my grans now. My train back to college is in two hours but I must leave for the station in one.

I didn't sleep very well last night; four out of the six fellow travelers in my compartment happened to be heavy snorers.

Gosh, one of their snores sounded incredibly peculiar, as if a zip was being opened and closed at regular intervals.

@JohnRennie Hahaha! Oh, this reminds me that I must buy earplugs and an eye mask as soon as possible!

@BalarkaSen Dang, what are you having?

@JohnRennie Nope! Will do, in half an hour...

One egg?

@BalarkaSen I have googled "Viscera" and wow.

Just blindfold yourself and eat it. Only the taste matters anyway :P

Yo @Balarka: After you are finished with lunch, you think you can quickly help me to understand something?

Now?

OK, so I am s'posed to find the absolute maximum and minimum values of a function $f(x,y)$ over a given region $R$.

Here is a sample problem:

Agh, it is taking far too long to load, the image!

(Praise the Lord!)

@JohnRennie It's not only you. It'll probably also take me a lifetime(maybe more) to understand the essence of weird meme humor. I've stopped trying. :P

Right, my question is this: what do they mean when they say "We want to determine the locations of the points on the boundary of $R$ at which the absolute extrema might occur"?

First find the maxima/minima points and then determine which of them lie on the boundary

The maxima/minima/saddle points occur where the gradient is $0$

i.e. $\nabla f =0$

@BalarkaSen How come?

(@Blue: I see...)

Yes, I s'pose...

Alright, I don't fully understand this yet and I will, if I spend 7 more minutes with this, I'm sure, but I must move on; it will do to remember this:

Thanks! :-)

On the interior? That would be at $x=1$, no? It doesn't look like this function would have an absolute maxima...

By equating its slope to zero, and solving for $x$.

Its slope is 1, because of which, erm, well, it's never 0, suggesting that it doesn't have a critical point, perhaps?