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?
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 $f(n)$.