Given $N$ and $a$ positive integers, with $a\ge 2$ is it possible to prove the inequality: $$\sum_{k=1}^N\frac{k^a}{(k+1)^a+(k+2)^a}\le\frac{N}{2}$$
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2$\begingroup$ 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 $\endgroup$– Dominic MichaelisCommented May 21, 2013 at 13:11

$\begingroup$ You can use the principle of mathematical induction to prove such inequalities. $\endgroup$– aslanCommented May 21, 2013 at 14:11
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1 Answer
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Since $\displaystyle \frac{k^a}{(k+1)^a+(k+2)^a}<\frac{k^a}{k^a+k^a}=\frac{1}{2}$ then $\displaystyle\sum_{k=1}^N\frac{k^a}{(k+1)^a+(k+2)^a}<\sum_{k=1}^N\frac{1}{2}=\frac{N}{2}$.