see:old post and Make the integral discrete, we have
conjecture :for any complex numbers $a_{i},i=1,2,\cdots,n$,and $p_{i}\ge 0$ such $p_{1}+p_{2}+\cdots+p_{n}=1$,then maybe be have $$\sum_{i=1}^{n}\sum_{j=1}^{n}|a_{i}+a_{j}|\cdot p_{i}p_{j}\ge \sum_{i=1}^{n}p_{i}|a_{i}|$$
How prove this complex inequality?Thanks
take $p_{i}=\dfrac{1}{n},i=1,2,\cdots,n$,and Amuess that $a_{i}\in R$.I have prove that following
**Theorem **.$$\sum_{i,j=1}^{n}|a_{i}+a_{j}|\ge n\sum_{i=1}^{n}|a_{i}|$$ proof: Equality occurs if all $x_k$ are null, or else if $n$ is even, half the variables are equal to any real number $a$, and the other half are equal to $-a$. The proof goes along the lines exposed in the above, but an alternate solution, given by C. Tanasescu, is worth mentioning.
Use the obvious equality $|a| + |b| - |a+b| = 0$ if $ab\geq 0$ and $= 2\min\{|a|,|b|\}$ if $ab<0$. Then $$\begin{align*}\sum_{1\leq i,j\leq n} (|x_i| + |x_j| - |x_i+x_j|) &= \sum_{x_ix_j< 0} 2\min\{|x_i|,|x_j|\}\\ & = 4\sum_{x_i>0>x_j}\min\{x_i,-x_j\} \leq 4\sum_{x_i>0>x_j}\sqrt{-x_ix_j} \\ &= 4\sum_{x_i>0}\sqrt{x_i} \sum_{0>x_j}\sqrt{-x_j} \leq \left (\sum_{x_i>0}\sqrt{x_i} + \sum_{0>x_j}\sqrt{-x_j}\right )^2 \\ &=\left (\sum_{1\leq k\leq n}\sqrt{|x_k|}\right )^2 \\ &\leq n\sum_{1\leq k\leq n}|x_k|\end{align*}$$
It follows $\sum_{1\leq i,j\leq n} |x_i+x_j| + n\sum_{1\leq k\leq n}|x_k| \geq \sum_{1\leq i,j\leq n} (|x_i| + |x_j|) = 2n\sum_{1\leq k\leq n}|x_k|$, which is equivalent to the required inequality
