Can you help me realize why this is true? I can tell you that $P_i$ and $P_j$ are probabilities, i.e. $0 \leq P_i, P_j \leq 1$.
$\displaystyle \sum_{i=1}^\infty \sum_{j=1}^\infty ijP_iP_j \leq \sum_{i=1}^\infty \sum_{j=1}^\infty j^2P_jP_i$.
Can you help me realize why this is true? I can tell you that $P_i$ and $P_j$ are probabilities, i.e. $0 \leq P_i, P_j \leq 1$.
$\displaystyle \sum_{i=1}^\infty \sum_{j=1}^\infty ijP_iP_j \leq \sum_{i=1}^\infty \sum_{j=1}^\infty j^2P_jP_i$.
For a given pair $(m,n)$ with $m\ne n$, consider the sum of the two terms corresponding to $(i,j) = (m,n)$ and $(i,j)=(n,m)$. For the summation on the left, this will be $$ mnP_mP_n + nmP_nP_m = 2mnP_mP_n, $$ while for the sum on the right this will be $$ n^2P_n P_m + m^2 P_m P_n = (n^2+m^2)P_mP_n. $$ Since $2mn \le m^2+n^2$ for all $m$ and $n$, it follows that $$ \sum_{i \ne j} ijP_iP_j \le \sum_{i\ne j} j^2 P_j P_i. $$ But $$ \sum_{i = j} ijP_iP_j = \sum_{i = j} j^2 P_j P_i, $$ so this proves the given inequality.