BdMO Online Forum

Prove that \[\sum_{k=1}^n \sqrt{k^2+1}\geq n\sqrt{\dfrac{n^2+2n+5}{4}}\] holds for all \(n\in\mathbb{N}\).

Straightforward Jensen...

Let, $f(x)=\sqrt{1+x^2}\Rightarrow f''(x)=(1+x^2)^{-\frac{3}{2}}> 0 \; \forall x\in \mathbb{R}$
$\therefore f(x)$ is convex.
So,
\[\frac{\sum_{k=1}^n f(k)}{n}\geq f\left(\frac{\sum_{k=1}^n k}{n}\right)\]
\[\Rightarrow \sum_{k=1}^n\sqrt{k^2+1}\geq n\cdot f\left(\frac{n+1}{2}\right)\]
\[\Rightarrow \sum_{k=1}^n\sqrt{k^2+1}\geq n \sqrt{1+\frac{(n+1)^2}{4}}\]
\[\Rightarrow \sum_{k=1}^n \sqrt{k^2+1}\geq n\sqrt{\dfrac{n^2+2n+5}{4}}\]

Yes that was trivial by Jensen. Another way to see that \(f\) is convex is noticing that \(f\) is even and \(\lim f= |x|\). So the graph looks similar to \(y=|x|\) with a little curve.

A direct application of this.

Care to elaborate?