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Prove that inequality $$\sum_{k=1}^n (2k-1)^m \geq n^{m+1}$$ holds if $n,m$ are integer positive numbers.

Karamata`s inequality doesn`t work.

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  • $\begingroup$ What about taking a $log_{n}$ from both sides? $\endgroup$ – openspace May 15 '16 at 19:11
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We can do it by induction on $m$. For $m=1$ easy to check it's actually equality.

Suppose the inequality is true for $m$, then for $m+1$ we have

$$\sum_{k=1}^n (2k-1)^{m+1}=\sum_{k=1}^n (2k-1)^m(2k-1)\geq \frac{1}{n}(\sum_{k=1}^n (2k-1)^{m})(\sum_{k=1}^n (2k-1))\geq \frac{1}{n}n^{m+1}n^2=n^{m+2}$$

First inequality by Chebyshev's sum inequality and second inequality by inductive assumption.

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It can be done without induction. Suppose that $0<a<1$. Then

$$\begin{align*} (1-a)^m+(1+a)^m&=\sum_{k=0}^m\binom{m}k(-1)^ka^k+\sum_{k=0}^m\binom{m}ka^k\\ &=\sum_{k=0}^m\binom{m}ka^k\big(1+(-1)^k\big)\\ &=1+\sum_{k=1}^m\binom{m}ka^k\big(1+(-1)^k\big)\\ &\ge 1 \end{align*}$$

for all $m\in\Bbb Z^+$. Thus,

$$\begin{align*} \sum_{k=1}^n(2k-1)^m&=n^m\sum_{k=1}^n\left(\frac{2k-1}n\right)^m\\ &=\begin{cases} n^m\sum_{k=1}^{n/2}\left(\left(1-\frac{2k-1}n\right)^m+\left(1+\frac{2k-1}n\right)^m\right),&\text{if }n\text{ is even}\\ n^m\left(1+\sum_{k=1}^{(n-1)/2}\left(\left(1-\frac{2k}n\right)^m+\left(1+\frac{2k}n\right)^m\right)\right),&\text{if }n\text{ is odd} \end{cases}\\ &\ge n\cdot n^m\\ &=n^{m+1}\;. \end{align*}$$

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By induction it is easy to see that $\sum_{k=1}^n (2k-1)^m \geq n^{m+1}$ is true for m=1. Now suppose that $$\sum_{k=1}^n (2k-1)^m \geq n^{m+1}$$ and let us prove it for $m+1$.

we have $$\sum_{k=1}^n (2k-1)^{m+1}=\sum_{k=1}^n (2k-1)^{m}(2k-1)\geq \frac{1}{n}\sum_{k=1}^n (2k-1)^{m}\sum_{k=1}^n(2k-1)\\ \geq \frac{1}{n} (n^{m+1})(2\frac{n}{2}(n+1)-n)=n^{m+2}$$ whish s the required.

Ps I have used the chebyshev inequality

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$ x\mapsto x^m $ is a convex function, hence $(n/2-\tau)^m+(n/2+\tau)^m \geq 2 n^m $.

Just sum these inequalities for appropriate values of $\tau$.

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