b 1 & lt； = b2 & lt=…& lt； =bn

(ai, bi > is not specified here; 0 )

Prove by mathematical induction:

When n=2, a 1b2+a2b 1

If n = k, it holds.

For disordered sum and ≤ inverse sum, N=k+ 1,

Take out those related to a 1 and b 1.

There is a 1bl+b 1at (out of order).

Give a 1bl+b 1at.

The remaining k terms satisfy the hypothesis.

For inverse sum and ≤ disordered sum

Take out those related to a 1, b(k+ 1).

At, there is a 1bl+b(k+ 1) (out of order).

Give a1bl+b (k+1) at >; = a 1b(k+ 1)+atbl & lt； = = & gt(a 1-at)(b(k+ 1)-bl)& lt； =0 hold

The remaining k terms satisfy the hypothesis.

In fact, the ranking inequality mainly applies to

& amp&&&&&& & (A-B) (C-D) > = 0&&&&& & &

I hope it helps you!