9.1. Let $a<b<c<d$ be real numbers. Arrange in ascending order the sums $a b+c d, a c+b d, a d+b c$.
Answer. $a d+b c, a c+b d, a b+c d$.

Solution. The difference between the first and second numbers from the condition is $a b+c d-(a c+b d)=d(c-b)+a(b-c)=(d-a)(c-b)>0$. Similarly, the difference between the second and third numbers from the condition is $a c+b d-(a d+b c)=b(d-c)+a(c-d)=(b-a)(d-c)>0$. Therefore, in the condition, these numbers are arranged in descending order, and in the answer, this order should be reversed.

Grading criteria. Only one pair of numbers is correctly ordered with justification: 3 points.