The sum of $m$th powers of positive odd integers up to $2n-1$ is at least $n^{m+1}$ 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.
 A: 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.
A: 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,
$$\begin{align*}
\sum_{k=1}^n(2k-1)^m&=n^m\sum_{k=1}^n\left(\frac{2k-1}n\right)^m\\
&=\begin{cases}
n^m\sum_{k=1}^{n/2}\left(\left(1-\frac{2k-1}n\right)^m+\left(1+\frac{2k-1}n\right)^m\right),&\text{if }n\text{ is even}\\
n^m\left(1+\sum_{k=1}^{(n-1)/2}\left(\left(1-\frac{2k}n\right)^m+\left(1+\frac{2k}n\right)^m\right)\right),&\text{if }n\text{ is odd}
\end{cases}\\
&\ge n\cdot n^m\\
&=n^{m+1}\;.
\end{align*}$$
A: 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
A: $ 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 $\tau$.
