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

Question

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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

Well an obivous bound for the left hand side is $1006 \cdot 1007$ try to get it sharper — Dominic Michaelis, Feb 23 '13 at 20:03
What does that even mean XD I'm not really good at mathematics, hence I'm asking it online :D — mathsnoob, Feb 23 '13 at 20:07
nobody is really good at Maths, we all try, so try something too. — , Feb 23 '13 at 20:12
Ouch... I have tried, but I really don't have a way with inequalities, so maybe help me out a bit? :) — mathsnoob, Feb 23 '13 at 20:15
"Young man, in mathematics you don't understand things. You just get used to them." John von Neumann. — Julien, 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
@HaraldHanche-Olsen my thought exactly. looks like a contest's problem to me but one can't say if it is current or not. — , Feb 23 '13 at 20:20
It's not for a contest, it's my maths homework for practising for IMO (which is a competition) — mathsnoob, Feb 23 '13 at 20:21
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
A possible generalization: Prove the inequality $\sum_{k=1}^{2n-1}\sqrt{k(4n-k)}<\pi n^2$ for all natural $n$ — Martin Sleziak, Jan 05 '20 at 11:41

score 6 · Accepted Answer · answered Feb 23 '13 at 20:20

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

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

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