Take the tour ×  Given $N$ and $a$ positive integers, with $a\ge
2$ is it possible to prove the inequality:
$$\sum_{k=1}^N\frac{k^a}{(k+1)^a+(k+2)^a}\le\frac{N}{2}$$
share|improve this question I don't know if this one is
appropiate on this site. As $\frac{1}{N} \cdot \sum_{k=1}^N
\dots $ converges to $\frac{1}{2}$ you could show that it is
monotone increasing and you are done –  Dominic Michaelis May 21
at 13:11 You can use the principle of mathematical induction to
prove such inequalities. –  JD Vlok May 21 at 14:11 add comment
1 Answer  up vote 2 down vote accepted  Since $\displaystyle
\frac{k^a}{(k+1)^a+(k+2)^a}<\frac{k^a}{k^a+k^a}=\frac{1}{2}$
then $\displaystyle\sum_{k=1}^N\frac{k^a}{(k+1)^a+(k+2)^a}<\sum_
{k=1}^N\frac{1}{2}=\frac{N}{2}$.  share|improve this answer add
comment  Your Answer