How do we prove for any given real number $a_1,\ldots,a_n$, we have $$\frac{1}{n^2}\sum_{i=1}^n\sum_{j=1}^n|a_i+a_j|\geq \frac{1}{n}\sum_{i=1}^n|a_i|$$
I have checked several easy cases for this:
When $n=1$, LHS $=2|a_1|$ and RHS $=|a_1|$, so it trivially holds.
When $n=2$, we are going to prove $$2|a_1|+2|a_1+a_2|+2|a_2|\geq 2(|a_1|+|a_2|)\Longleftrightarrow 2|a_1+a_2|\geq 0$$ which is also trivial.
When $n=3$, we are going to prove $$\sum_{i=1}^3\sum_{j=1}^3|a_i+a_j|\geq 3\sum_{i=1}^n|a_i|$$ which is equivalent to prove $$2(|a_1+a_2|+|a_2+a_3|+|a_3+a_1|)\geq |a_1|+|a_2|+|a_3|$$ This can be proved simply by triangular inequality, and actually the coefficient $2$ is not sharp, $1.5$ is enough.
However, when $n$ gets larger, e.g., $n\geq 6$, simply applying triangular inequality is not enough because by in general it reduces to prove: $$2\sum_{i<j}|a_i+a_j|\geq (n-2)\sum_{i=1}^n|a_i|$$
I don't know how to proceed, can anyone help me?
Edit: Since this problem is pointed out to be duplicate, I slightly generalize it. If $b_1,\ldots,b_n$ are convex combination coefficient, i.e., they are nonnegative and $b_1+\ldots+b_n=1$, then how to prove $$\sum_{i=1}^n\sum_{j=1}^nb_ib_j|a_i+a_j|\geq \sum_{i=1}^nb_i|a_i|$$
