<s>[INST] Help with the inequality $\sum_{k=1}^{1006} \sqrt{k \cdot (2014-k)}<506^2\pi$ 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$$\pi$
Thanks
 [/INST] $$\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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