aj-ai & gt； = {( 1+ai)( 1+aj)}/ 10 10
Namely10/10 10 >；; =( 1+ai)( 1+aj)/(aj-ai)
(aj-ai)/(( 1+ai)( 1+aj))& gt； = 1/ 10 10
(( 1+aj)-( 1+ai))/(( 1+ai)( 1+aj))& gt； = 1/ 10 10
That is,1(1+ai)-1(1+aj) > =110.
Because for any AI, AJ, (j >；; I) The above formula holds, so there must be:
1/( 1+a 1)- 1/( 1+a2)>= 1/ 10 10
1/( 1+a2)- 1/( 1+a3)>= 1/ 10 10
......
1/( 1+a 20 10)- 1/( 1+a 20 10)>； = 1/ 10 10
***20 10 formula, two sides are added, and the direction of the unequal sign is unchanged, in which:
1/( 1+a 1)- 1/( 1+a 20 10)>= 1
1/( 1+a 1)>= 1/( 1+a20 10)>+ 1 & gt； 1
Therefore: 1+A 1
That is a 1 < 0.
This is similar to a1>; 0 is contradictory, so the original assumption is untenable, which means that there must be ai, aj, (I
aj-ai & lt； {( 1+ai)( 1+aj)}/ 10 10