I am sure this question will have been asked somewhere but I can't find it anywhere. Let $X=S_1+S_2$ where $S_1$ and $S_2$ are the outcomes of two independently rolled biased die(1 to 6), where $P(S_1=i)=p_i$ and $P(S_2=i)=q_i$. I am asked to prove that, by considering $P(X=7)$ and $\sqrt{P(X=2)P(X=12)}$ using the AM-GM inequality it is not possible to find $p_i$ and $q_i$ such that $P(X=2)=P(X=3)=...+P(X=12)$.
I have already solved this problem in a different way, I just assumed they were equal then squared $P(X=7)=(p_1q_6+...+p_6q_1)$ and if you expand things out in a clever way you can find $p_1q_6p_2q_6$ in there which equals $P(X=2)P(X=12)$, and you can then conclude all the probabilities equal zero since $P(X=7)^2=P(X=2)P(X=12)$ under the assumption. However, I am struggling to even start using AMGM. I suppose I would go something like $\sqrt{P(X=2)P(X=12)}\le\frac{P(X=2)+P(X=12)}{2}$, however, we know that there is equality if and only if $P(X=12)=P(X=2)$ as a property of the AMGM so I don't know how what we get can possibly contradict that.
