Proofathon Inequality

For discussing Olympiad Level Algebra (and Inequality) problems
Nirjhor
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Proofathon Inequality

Unread post by Nirjhor » Thu Aug 21, 2014 2:09 am

Prove that \[\sum_{k=1}^n \sqrt{k^2+1}\geq n\sqrt{\dfrac{n^2+2n+5}{4}}\] holds for all \(n\in\mathbb{N}\).
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.


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sowmitra
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Re: Proofathon Inequality

Unread post by sowmitra » Thu Aug 21, 2014 6:40 pm

Straightforward Jensen...

Let, $f(x)=\sqrt{1+x^2}\Rightarrow f''(x)=(1+x^2)^{-\frac{3}{2}}> 0 \; \forall x\in \mathbb{R}$
$\therefore f(x)$ is convex.
So,
\[\frac{\sum_{k=1}^n f(k)}{n}\geq f\left(\frac{\sum_{k=1}^n k}{n}\right)\]
\[\Rightarrow \sum_{k=1}^n\sqrt{k^2+1}\geq n\cdot f\left(\frac{n+1}{2}\right)\]
\[\Rightarrow \sum_{k=1}^n\sqrt{k^2+1}\geq n \sqrt{1+\frac{(n+1)^2}{4}}\]
\[\Rightarrow \sum_{k=1}^n \sqrt{k^2+1}\geq n\sqrt{\dfrac{n^2+2n+5}{4}}\]
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Nirjhor
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Joined:Thu Aug 29, 2013 11:21 pm
Location:Varies.

Re: Proofathon Inequality

Unread post by Nirjhor » Thu Aug 21, 2014 9:11 pm

Yes that was trivial by Jensen. Another way to see that \(f\) is convex is noticing that \(f\) is even and \(\lim f= |x|\). So the graph looks similar to \(y=|x|\) with a little curve. :)
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.


Revive the IMO marathon.

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Mursalin
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Joined:Thu Aug 22, 2013 9:11 pm
Location:Dhaka, Bangladesh.

Re: Proofathon Inequality

Unread post by Mursalin » Fri Aug 22, 2014 6:37 pm

A direct application of this.

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Nirjhor
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Re: Proofathon Inequality

Unread post by Nirjhor » Fri Aug 22, 2014 7:24 pm

Care to elaborate? ;)
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.


Revive the IMO marathon.

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