An Application of Cauchy Schwarz - AM-GM in Discrete Probability Measures Suppose $p_m \geq 0$ and $\sum_{m \in \mathbf{Z}} p_m =1 .$ That is $p$ is a probability measure on integers. Then how can I show (is it true) that
$$ \sum_{m \in \mathbf{Z}} (p_m + p_{m+1} + p_{m+2})^n \leq 2^n \sum_{m \in \mathbf{Z}} (p_m + p_{m+1})^n - (2^n-1) \sum_{m \in \mathbf{Z}} p_m ^n  $$
for any $n \in \mathbf{N}.$
If I am on not too far, PLEASE do NOT answer, yet provide me a hint (I do not want to ruin the joy). Yet I have been stuck for a bit, thank you!

I tried to use cauchy schwarz, and I believe some form of such an inequality should work. As a simple case, consider $n=2:$ After changing indices, we have
\begin{align}
\textbf{Left} = \sum_{m \in \mathbf{Z}} (p_m + p_{m+1} + p_{m+2})^2 =3 \sum_{m \in \mathbf{Z}} p_m ^2 + 4 \sum_{m \in \mathbf{Z}} p_m p_{m+1} + 2 \sum_{m \in \mathbf{Z}} p_m p_{m+2} 
\end{align}
and
\begin{align}
\textbf{Right} = 2^2 \sum_{m \in \mathbf{Z}} (p_m + p_{m+1})^2 - (2^2-1) \sum_{m \in \mathbf{Z}} p_m ^2 &= 4 \Big[ 2 \sum_{m \in \mathbf{Z}} p_m ^2 + 2 \sum_{m \in \mathbf{Z}} p_m p_{m+1} \Big] - 3 \sum_{m \in \mathbf{Z}} p_m ^2 \\
&= 5 \sum_{m \in \mathbf{Z}} p_m ^2 + 8 \sum_{m \in \mathbf{Z}} p_m p_{m+1} .
\end{align}
By cauchy schwarz,
\begin{align}
\sum_{m \in \mathbf{Z}} p_m p_{m+2} \leq \Big( \sum_{m \in \mathbf{Z}} p_m ^2 \Big) ^{\frac{1}{2}} \Big( \sum_{m \in \mathbf{Z}} p_{m+2} ^2 \Big) ^{\frac{1}{2}} = \sum_{m \in \mathbf{Z}} p_m ^2  
\end{align}
which suffices to show $\textbf{Left} \leq \textbf{Right}.$ As another way, by AM-GM we have
\begin{align}
\sum_{m \in \mathbf{Z}} p_m ^2 = \frac{1}{2} \sum_{m \in \mathbf{Z}} p_m ^2 + \frac{1}{2} \sum_{m \in \mathbf{Z}} p_{m+2} ^2 = \sum_{m \in \mathbf{Z}}  \frac{p_m ^2 + p_{m+2} ^2 }{2}  \geq \sum_{m \in \mathbf{Z}} \sqrt{p_m ^2  p_{m+2} ^2 } =  \sum_{m \in \mathbf{Z}} p_m p_{m+2} .
\end{align}
For $n=3 :$
After changing indices, we have
\begin{align}
\textbf{Left} = \sum_{m \in \mathbf{Z}} (p_m + p_{m+1} + p_{m+2})^3 &=3 \sum_{m \in \mathbf{Z}} p_m ^3 + 6 \sum_{m \in \mathbf{Z}} p_m^2 p_{m+1} + 3 \sum_{m \in \mathbf{Z}} p_m^2 p_{m+2} \\
&+ 6 \sum_{m \in \mathbf{Z}} p_m p_{m+1} ^2 + 3 \sum_{m \in \mathbf{Z}} p_m p_{m+2} ^2 + 6 \sum_{m \in \mathbf{Z}} p_m p_{m+1} p_{m+2} 
\end{align}
and
\begin{align}
\textbf{Right} = 2^3 \sum_{m \in \mathbf{Z}} (p_m + p_{m+1})^3 - (2^3-1) \sum_{m \in \mathbf{Z}} p_m ^3 &= 8 \Big[ 2 \sum_{m \in \mathbf{Z}} p_m ^3 + 3 \sum_{m \in \mathbf{Z}} p_m ^2 p_{m+1} + 3 \sum_{m \in \mathbf{Z}} p_m p_{m+1} ^2 \Big] - 7 \sum_{m \in \mathbf{Z}} p_m ^3 \\
&= 9 \sum_{m \in \mathbf{Z}} p_m ^3 + 24 \sum_{m \in \mathbf{Z}} p_m ^2 p_{m+1} + 24 \sum_{m \in \mathbf{Z}} p_m p_{m+1} ^2 .
\end{align}
One can perform cancelations and similarly show by cauchy schwarz or AM-GM
\begin{align}
\sum_{m \in \mathbf{Z}} p_m ^2 p_{m+2} \leq \sum_{m \in \mathbf{Z}} p_m ^3  
\end{align}
and
\begin{align}
\sum_{m \in \mathbf{Z}} p_m p_{m+2} ^2 \leq \sum_{m \in \mathbf{Z}} p_m ^3  
\end{align}
and using AM-GM
\begin{align}
\sum_{m \in \mathbf{Z}} p_m ^2 p_{m+1} + \sum_{m \in \mathbf{Z}} p_m p_{m+1} ^2 = \sum_{m \in \mathbf{Z}} \frac{p_m ^2 p_{m+1} + p_{m+1} p_{m+2}^2 }{2} \geq \sum_{m \in \mathbf{Z}} p_m p_{m+1} p_{m+2} .
\end{align}
