0
$\begingroup$

Another inequality with powers: The proof for previous inequality does not trivially extend, I guess.

$\text{For}\; n>2,\quad\quad(2n)^{n-1}> 1^{n-1}+3^{n-1}+...+\left( 2n-3\right) ^{n-1}+\left( 2n-1\right) ^{n-1}$

  • 0
    You are absolutely right, corrected.2012-11-26

1 Answers 1

0

If we divide both sides by $2^{n-1}$, we have to prove: $\forall n>2,\quad n^{n-1} > \sum_{k=0}^{n-1}\left(k+\frac{1}{2}\right)^{n-1}.$ Since $f(x)=x^{n-1}$ is a positive, continous, increasing and (midpoint-)convex function on $\mathbb{R}^+$, we have: $ \left(k+\frac{1}{2}\right)^{n-1}< \int_{k}^{k+1}f(x)\,dx,$ so: $ \sum_{k=0}^{n-1}\left(k+\frac{1}{2}\right)^{n-1}<\int_{0}^{n}x^{n-1}\,dx=n^{n-1},$ QED.

  • 0
    Excellent, thanks.2012-11-27