The last time I post one of these, I got some very creative and excellent responses. Here's another question. This one requires some knowledge of calculus. For 10kpp,

Prove:

For real-valued functions f, and g,

Int_0^1 Int_0^1 |f(x)g(y) + f(y)g(x)|dxdy >= (Int_0^1 |f(x)|dx)(Int_0^1 |g(y)|dy).

where Int_a^b := Integral from a to b,
and a >= b means "a greater or equal to b."

Looks like hardcore 1337-speak. :(

But, sorry...I don't know Calculus yet.

I've never seen anyone offer someone else video game currency to do their assignments before...

I've never seen anyone offer someone else video game currency to do their assignments before...
Maybe you haven't been paying much attention.

http://forums.ayonae.ro/showthread.php?t=324

{I predict Palarran wins within 3 days}

I've never seen anyone offer someone else video game currency to do their assignments before...
:p :) :p

lol Sanchek, you actually have that old post. Well, this question is different (not a homework assignment either...would seriously kill a student if you give to them for no reason). I was reading a paper where the authors mentioned that this is clearly true and I have not been able to see this obvious statement. There are quite some very smart people around here so I wish they can have fun with this. Again, for the most, if you want to make some hourly wage in terms of pp, I would recommend you goto rathe mountain and kill giants... you will be much better off with that! :D

Palarran straight up owns math. I wasn't joking about that!

Heh, I've been struggling with this problem. The result makes sense intuitively but as far as proving it goes I'm still missing something.

I wish I was mathematically intelligent, as I'm flat broke :(

lol, hell i failed Pre-Algebra, got a 'D' in Algebra, and failed Geometery.

I think the answer is 5

Bah - Its 42!

Lani

Bah - Its 42!

Lani
But what was the question? :)

What's 6x7?

....

Invalid, Valid. You can't just do the simple squaring when you integrate because they are arbitrary real functions. f(x) could be e^x, which doesn't integrate like you suggest.

I've only given a little more thought besides my original special case solution (posted in some other thread). I think it might be fruitful to use Cauchy's inequality (a.k.a. Triangle inequality) in some clever way. It says that |a| - |b| <= |a+b| <= |a| + |b|. Haven't given it too much consideration though.

....

answer is greater than, or less than, or equal to zero.

I'm not wrong!

For one thing, you went from:
Int 0^1 Int 0^1 | f(x)g(y) + f(y)g(x) | dxdy
directly to:
[F(1) g(y)] + [G(1) f(y))] - [F(0) g(y)] - [G(0) f(y)]

dropping the absolute value sign somehow.

Oh, and suppose the answer is something like i...then the answer is not greater than, less than, or equal to zero, since the set of complex numbers is not ordered. Is i>0? It's meaningless! :P

The original problem statement said the functions were real valued. But still, no dealing with the ABS.

I was referring to Liper who claimed that the "answer" was greater than, less than, or equal to zero. :)

This one seems to be a bit more of a stumper than the previous one. The first half of that one, anyway =P

I miss being able to do problems similar to these. Math seems so much easier when you're doing it every day as a matter of course. Been 3 years since I've had to solve anything and I doubt I figure out tall a flagpole is from the length of the shadow. 8/