# Help with the inequality $\sum_{k=1}^{1006} \sqrt{k \cdot (2014-k)}<506^2\pi$

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