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PostPosted: Mon, 14 Jan 2008 10:23:45 UTC 
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Challenge other members if you have interesting problems in Inequalities or Integration!

You do not need to know the solution yourself!

Please do NOT post easy, boring, or IRRELEVANT problems in this thread!

I will post my problems later, but first I'd like to see some activities from other members! Let's see what you've got! :)


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 Post subject: Inequalities 1, 2
PostPosted: Mon, 14 Jan 2008 14:27:30 UTC 
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Hello. Here is an inequality(1) (not proved by me yet, but I think it's proved by someone else, i.e. it's not a conjecture).

Prove that for all n \in \mathbb{N} the following holds:
\sqrt[n]{\dbinom{2n+1}{n}} \geq 2\left(1+\frac{1}{\sqrt[n]{n+1}}\right)


Another inequality (2). (This one is known)
Let a, b and c be positive reals. Prove that

\displaystyle \sqrt{\frac{a}{a+b}} + \sqrt{\frac{b}{b+c}} + \sqrt{\frac{c}{c+a}} \leq \frac{3\sqrt{2}}{2}


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 Post subject: Re: Inequalities 1, 2
PostPosted: Tue, 15 Jan 2008 17:08:16 UTC 
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kar wrote:
Prove that for all n \in \mathbb{N} the following holds:
\sqrt[n]{\dbinom{2n+1}{n}} \geq 2\left(1+\frac{1}{\sqrt[n]{n+1}}\right)

this is a very strange result! i've found a few lower bounds for $ \binom{2n+1}{n} but all of them are annoyingly smaller
than $ 2^n\left(1+\frac{1}{\sqrt[n]{n+1}}\right)^n. i used a strong version of Stirling's approximation but that made things even more

complicated. i have no idea how to prove this! anyone?


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PostPosted: Sun, 27 Jan 2008 18:32:13 UTC 
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Problem 1 Easy! I've mentioned it earlier in Algebra forum:

Show that for all x \in \mathbb{R}: \ x^{2n}+2x^{2n-1}+3x^{2n-2}+ ... + 2n +1  \geq n+1. Find all x for which equality holds.

Problem 2 Hard I guess, and I don't know the answer:
For a given natural number n and integers (or more generally reals) 0< a_0 < a_1 < ... < a_{2n}, find $ \min_{x \in \mathbb{R}}\sum_{j=0}^{2n}a_jx^{2n-j}.


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PostPosted: Tue, 29 Jan 2008 09:03:38 UTC 
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This is supposedly made by me (I don't remember).
suppose a,b,c are positive reals such that abc=1
show:
\frac{1}{\sqrt{(1+a)(1+b)}} + \frac{1}{\sqrt{(1+a)(1+c)}} + \frac{1}{\sqrt{(1+b)(1+c)}} \leq \frac{3}{2}

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Has anyone noticed that the below is WRONG? Otherwise this statement would be true:
-1\cong1\pmod{13}
i\cong5 \pmod{13} where
i^2=-1


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PostPosted: Mon, 4 Feb 2008 12:36:39 UTC 
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Try this problem:
Prove that \displaystyle\int_0^{\frac{\pi}{2}}\ln(\sin\theta) d\theta =-\frac{\pi}{2}\ln (2).

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PostPosted: Tue, 5 Feb 2008 03:48:54 UTC 
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bidentate wrote:
Try this problem: Prove that \displaystyle\int_0^{\frac{\pi}{2}}\ln(\sin\theta) d\theta =-\frac{\pi}{2}\ln (2).

it's quite easy: let $ I=\int_0^{\frac{\pi}{2}}\ln \sin\theta \ d\theta. then $ I=\int_0^{\frac{\pi}{2}}\ln 2 \ d\theta + \int_0^{\frac{\pi}{2}}\ln \sin(\frac{\theta}{2}) d\theta + \int_0^{\frac{\pi}{2}}\ln \cos(\frac{\theta}{2}) d\theta
$ =\frac{\pi}{2}\ln 2 + \int_0^{\frac{\pi}{2}}\ln \sin(\frac{\theta}{2}) d\theta + \int_{\frac{\pi}{2}}^{\pi}\ln \sin(\frac{\theta}{2}) d\theta= \frac{\pi}{2}\ln 2 + \int_0^{\pi}\ln \sin(\frac{\theta}{2}) d\theta = \frac{\pi}{2}\ln 2 + 2I. therefore:
$ I - 2I = \frac{\pi}{2}\ln 2, and hence $ I=-\frac{\pi}{2}\ln 2.


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PostPosted: Wed, 6 Feb 2008 09:20:57 UTC 
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Prove, by induction, that for all integers $ n \geq 3: \ (n^2-1)! > n^{n^2}.

Note: For a non-inductive proof of this inequality see this thread.


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PostPosted: Mon, 11 Feb 2008 20:28:02 UTC 
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How about a problem that is both an inequality and an integral?

Let f(x) be a function such that \int_0^1 f(x)dx=1.

Prove that \int_0^1 (1-f(x))e^{-f(x)}dx \geq 0 and only equals 0 when f(x)=1 almost everywhere.


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PostPosted: Mon, 11 Feb 2008 22:26:35 UTC 
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Zathras wrote:
How about a problem that is both an inequality and an integral?

Let f(x) be a function such that \int_0^1 f(x)dx=1.

Prove that \int_0^1 (1-f(x))e^{-f(x)}dx \geq 0 and only equals 0 when f(x)=1 almost everywhere.


By almost everywhere, do you mean off a set of measure 0?

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 Post subject:
PostPosted: Mon, 11 Feb 2008 23:24:35 UTC 
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Shadow wrote:
Zathras wrote:
How about a problem that is both an inequality and an integral?

Let f(x) be a function such that \int_0^1 f(x)dx=1.

Prove that \int_0^1 (1-f(x))e^{-f(x)}dx \geq 0 and only equals 0 when f(x)=1 almost everywhere.


By almost everywhere, do you mean off a set of measure 0?


Yes.


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PostPosted: Tue, 12 Feb 2008 01:29:40 UTC 
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Zathras wrote:
How about a problem that is both an inequality and an integral?

Let f(x) be a function such that \int_0^1 f(x)dx=1.

Prove that \int_0^1 (1-f(x))e^{-f(x)}dx \geq 0 and only equals 0 when f(x)=1 almost everywhere.

let 1-f(x)=g(x). then \int_0^1 g(x) dx = 0, and thus \int_0^1(1-f(x))e^{-f(x)}dx=e^{-1}\int_0^1 g(x)(e^{g(x)}-1)dx \geq 0,

because a(e^a - 1) \geq 0, for all a \in \mathbb{R}. the second part of your problem also follows quickly: since g(x)(e^{g(x)}-1) \geq 0,

for all x, we have \int_0^1 g(x)(e^{g(x)}-1)dx=0, iff g(x)=0 almost everywhere iff f(x)=1 almost everywhere. Q.E.D.


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PostPosted: Fri, 15 Feb 2008 09:10:24 UTC 
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For any integrable function f: [a,b]$ \longrightarrow \mathbb{R}: $ \int_a^b f(x)e^{f(x)} dx \geq \frac{1}{b-a}\int_a^b f(x) dx \int_a^b e^{f(x)} dx.


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PostPosted: Sun, 17 Feb 2008 02:38:31 UTC 
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PostPosted: Sun, 17 Feb 2008 21:43:51 UTC 
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dragade3789 wrote:
Image

Are the denominators of the 2nd and 3rd expression purposely equal or is it a mistake?


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