Prove that this sequence is decreasing While trying to prove some Hardy-type inequalities, I should claim that the following sequence is decreasing
$$
G(n) = f(n) \left[\frac{1}{n}\sum_{k = 1}^{n-1} f(k)\right]^{p-1}, 
$$
where the sequence $f(n)$ is decreasing and  $0<p<=1$.
I tried to prove that
$$
G(n+1) -G(n)<0
$$
as follows
$$
G(n+1)-G(n) =f(n+1)\left( \frac{1}{n+1}\sum_{k=1}^{n}f(k)\right)
^{p-1}-f(n)\left( \frac{1}{n}\sum_{k=1}^{n-1}f(k)\right) ^{p-1} \\
\leq f(n)\left( n+1\right) ^{1-p}\left( \sum_{k=1}^{n}f(k)\right)
^{p-1}-f(n)\left( n\right) ^{1-p}\left( \sum_{k=1}^{n-1}f(k)\right) ^{p-1},
$$
but could not estimate the right-hand side, Any help with this issue?
 A: Since $0<p\leq 1$, then
\begin{eqnarray*}
F(n+1)-F(n) &=&f(n+1)\left( \frac{1}{n+1}\sum_{k=1}^{n}f(k)\right)
^{p-1}-f(n)\left( \frac{1}{n}\sum_{k=1}^{n-1}f(k)\right) ^{p-1} \\
&=&f(n+1)\left( n+1\right) ^{1-p}\left( \sum_{k=1}^{n}f(k)\right)
^{p-1}-f(n)\left( n\right) ^{1-p}\left( \sum_{k=1}^{n-1}f(k)\right) ^{p-1},
\end{eqnarray*}
but since $f(n)$ is non-increasing (that is $f(n+1)\leq f(n)$) and $\left(
n\right) ^{1-p}\leq \left( n+1\right) ^{1-p}$, it follows that
\begin{equation}
F(n+1)-F(n)\leq f(n)\left( n+1\right) ^{1-p}\left( \sum_{k=1}^{n}f(k)\right)
^{p-1}-f(n)\left( n+1\right) ^{1-p}\left( \sum_{k=1}^{n-1}f(k)\right) ^{p-1},
\label{q}
\end{equation}
Now, since $f(n)\geq 0$
\begin{eqnarray*}
\sum_{k=1}^{n}f(k) &\equiv &f(1)+f(2)+\cdots +f(n-1)+f(n) \\
&=&\sum_{k=1}^{n-1}f(k)+f(n),
\end{eqnarray*}
then (note $p-1<0$)
\begin{equation*}
\left( \sum_{k=1}^{n}f(k)\right) ^{p-1}=\left(
f(n)+\sum_{k=1}^{n-1}f(k)\right) ^{p-1}\leq \left(
\sum_{k=1}^{n-1}f(k)\right) ^{p-1},
\end{equation*}
We finally get that
\begin{eqnarray*}
F(n+1)-F(n) &\leq &f(n)\left( n+1\right) ^{1-p}\left(
\sum_{k=1}^{n-1}f(k)\right) ^{p-1}-f(n)\left( n+1\right) ^{1-p}\left(
\sum_{k=1}^{n-1}f(k)\right) ^{p-1} \\
&=&0
\end{eqnarray*}
which asserts that $F(n)$ is also non-increasing as long as $f(n)$.
