Originally Posted by

**Showcase_22** I thought I could get there via an inequality route. =S

Discriminant= $\displaystyle 4n^3-4n^2(n)=4n^3-4n^3=0$

This implies that the quadratic only has 1 root and this root is repeated. Therefore a graph of this quadratic would have a minimum point that touches the x axis. Hence the entire quadratic would be greater than 0.

For a proof though, would it be better to construct the quadratic first, work all this out and eventually arrive at it must be greater than 0?