Proof.

If A' is unordered, there are two elements A'[i] > A'[i+1]. It is easy to show, that

A'[i] * B[i] + A'[i+1] * B[i+1] <= A'[i] * B[i+1] + A'[i] * B[i+1]

So we can swap elements A[i] and A[i+1] and the value of A'[1]*B[1]+..+A'[N]*B[N] will be the same or greater

We can repeat the process while A' is unordered

Hence, the greatest value of A'[1]*B[1]+..+A'[N]*B[N] is achieved when A' is non-decreasing

Elements of A and B can be negative.

Let A₁ < A₂ < ... A_n, and let i < j: A'_i > A'_j. Then we can swap A'_i and A'_j and sum A'₁·B₁ + ... + A'_n·B_n will increase. Then we will swap such pairs while its exist. And when there isn't such pair, array is sorted. Therefore, initial sum A'₁·B₁ + ... + A'_n·B_n less then sum A₁·B₁ + ... + A_n·B_n.

А зачем писать на английском?

Вот это неравенство в википедии

It's called the rearrangement inequality. Now that you know its name, I believe you can google it for more information?