# How to prove $\frac{n}{2} \sum_{i=1}^n a_i\,b_i \leqslant \sum_{i=1}^n a_i \sum_{i=1}^n b_i$ for $a_i,b_i\geq 0$?

I’m trying to prove if the following inequality holds, $$\frac{n}{2} \sum_{i=1}^n a_i\,b_i \leqslant \sum_{i=1}^n a_i \sum_{i=1}^n b_i,$$ where $a_i,b_i$ are real nonnegative numbers.

It looks like Chebyshev’s sum inequality but here I can’t make any assumption regarding the monotonicity of the two sequences $a_1,\ldots,a_n$ and $b_1,\ldots,b_n$, apart from the fact that are nonnegative.

Any help would be appreciated.

With $n=3$, $a_1=b_1=1$, and $a_2=a_3=b_2=b_3=0$, we have $$\frac{n}{2} \sum_{i=1}^n a_i b_i = \frac{3}{2} > 1 = \sum_{i=1}^n a_i \sum_{i=1}^n b_i$$