Prove the inequality $\sum_{k=1}^{2n-1}\sqrt{k(4n-k)}<\pi n^2$ for all natural $n$.
Please help me, I don't have an idea how to solve this.
Thank you.
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2Have you tried using an integral? What happens if you complete the square for $\sqrt{k(4n-k)}$? – abiessu Sep 08 '13 at 10:00
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1Have you considered proving it by induction? – Adriano Sep 08 '13 at 10:00
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@Adriano - the presence of the $n$ under the radix means that induction isn't trivial. – Glen O Sep 08 '13 at 10:15
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First, we can rewrite L.H.S of the inequality we want to prove as $$\begin{align} \sum_{k=1}^{2n-1}\sqrt{k(4n - k)} = & \sum_{k=1}^{2n-1}\sqrt{(2n)^2-(2n-k)^2} =\sum_{k=1}^{2n-1}\sqrt{(2n)^2-k^2}\\ = & 2n\sum_{k=1}^{2n-1}\sqrt{1-(k/2n)^2}\tag{*} \end{align}$$ Second, notice for $\displaystyle x \in \left[\frac{k-1}{2n},\frac{k}{2n}\right)$, we have $\displaystyle \sqrt{1-x^2} > \sqrt{1-(k/2n)^2}$, this implies: $$\int_{\frac{k-1}{2n}}^{\frac{k}{2n}} \sqrt{1-x^2} dx > \frac{1}{2n}\sqrt{1-(k/2n)^2}$$ Sum this up from $k = 1$ to $2n - 1$ and substitute this back into $(*)$, we get $$ \sum_{k=1}^{2n-1}\sqrt{k(4n - k)} < (2n)^2 \int_0^{1-\frac{1}{2n}}\sqrt{1-x^2}dx < (2n)^2 \int_0^1\sqrt{1-x^2}dx = \pi n^2 $$
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