# 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!

• Has this homework been asked explicitly under the form "prove using Cauchy-Schwarz inequality that ..." ? Or is it you who think that C.S. should be used ? Because at first sight, it is not striking that C.S. is a natural approach... May 29, 2022 at 7:33
• that is not a homework question, I wrote cauchy schwarz and AM-GM to name a few directions, I was checking < handouts from some olympiad notes to find useful things May 29, 2022 at 7:36
• A little remark: taking $n=1$ gives equality: LHS=RHS=3. May 29, 2022 at 7:43
• I just created a room called "PROBA". May 29, 2022 at 8:28
• chat.stackexchange.com/rooms/info/136671/proba?tab=general May 29, 2022 at 8:32

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}