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This question's been solved, come and look if you want to check out some hardcore solutions

Here's an inequality that needs to be proven:

Prove that

$\sqrt{1\cdot 2013} + \sqrt{2\cdot 2012} + \sqrt{3\cdot 2011} + \dots + \sqrt{1006\cdot 1008}$ < $506^2$$\pi$


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\cdot to get $\cdot$, \times to get $\times$. – Git Gud Feb 23 '13 at 19:59
nobody is really good at Maths, we all try, so try something too. – user31280 Feb 23 '13 at 20:12
"Young man, in mathematics you don't understand things. You just get used to them." John von Neumann. – 1015 Feb 23 '13 at 20:16
Since this is year 2013, this looks like a competition problem. Am I right? – Harald Hanche-Olsen Feb 23 '13 at 20:18
If so, I don't think @Marvis did you any favour by posting a complete answer. Just saying $\int\sqrt{x(1-x)}\,dx$ might have been better. Oh, well. – Harald Hanche-Olsen Feb 23 '13 at 20:23
up vote 6 down vote accepted

$$\sum_{k=0}^{n-1} \sqrt{k \cdot (2n-k)} = 2n \left(\sum_{k=0}^{n-1} \sqrt{\dfrac{k}{2n} \cdot \left(1-\dfrac{k}{2n} \right)} \right) = (2n)^2 \left(\sum_{k=0}^{n-1} \sqrt{\dfrac{k}{2n} \cdot \left(1-\dfrac{k}{2n} \right)} \cdot \dfrac1{2n}\right)$$ $$\underbrace{\left(\sum_{k=0}^{n-1} \sqrt{\dfrac{k}{2n} \cdot \left(1-\dfrac{k}{2n} \right)} \cdot \dfrac1{2n}\right) < \int_0^{1/2} \sqrt{x(1-x)}dx}_{\text{Since $\sqrt{x(1-x)}$ is a monotone increasing function for $x \in [0,1/2)$}} = \int_0^{\pi/4} 2\sin^2(y) \cos^2(y) dy = \dfrac{\pi}{16}$$ Hence, $$\sum_{k=1}^{n-1} \sqrt{k \cdot (2n-k)} < (2n)^2 \dfrac{\pi}{16} = \left(\dfrac{n}2 \right)^2 \pi$$

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Thanks! Life saviour – mathsnoob Feb 23 '13 at 20:27

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