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$n \cdot \left( \frac{n+1}{2} \right)^{\left( \frac{n+1}{2} \right)} \leqslant \sum_{k=1}^{n} k^k \text{ for } n \in \mathbb{N}$

I've tried to transform the left part of inequality, but nothing worked out: $$n \cdot \left( \frac{n+1}{2} \right)^{\left( \frac{n+1}{2} \right)} = \frac{n \cdot (n+1)^{\frac{n+1}{2}}}{2^{\frac{n+1}{2}}}$$ and then $$n \cdot (n+1)^{\frac{n+1}{2}} \leqslant 2^{\frac{n+1}{2}} \cdot \sum_{k=1}^{n} k^k$$

I will be very grateful if you give me some ideas how to transform this inequality in order to prove.

P.S There you can find what is Jensen's inequality.

Donald
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1 Answers1

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$$n \cdot \left( \frac{n+1}{2} \right)^{ \frac{n+1}{2} } \le \sum_{k=1}^{n} k^k$$

$$\left( \frac{n+1}{2} \right)^{ \frac{n+1}{2} } \le \frac{\sum_{k=1}^{n} k^k}n$$

$$\left( \frac{n(n+1)}{2n} \right)^{ \frac{n(n+1)}{2n} } \le \frac{\sum_{k=1}^{n} k^k}n$$

$$\left( \frac{1+2+…+n}{n} \right)^{ \frac{1+2+…+n}{n} } \le \frac{\sum_{k=1}^{n} k^k}n$$

This is just the Jensen’s inequality for the function $f(x)=x^x$ which is convex (= concave up).

We used the fact that $1+2+…+n=\frac{n(n+1)}2$.

Aig
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