Problem 1

Prove that the following assertion is true for

n=3

and

n=5

, and false for every other natural number

n>2

: for arbitrary real numbers

a_1,a_2,\ldots,a_n

(a_1-a_2)(a_1-a_3)\cdots(a_1-a_n)+(a_2-a_1)(a_2-a_3)\cdots(a_2-a_n)+\cdots+(a_n-a_1)(a_n-a_2)\cdots(a_n-a_{n-1})\ge0

Step 4 of 6: Group the first two terms for n=5

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(a_1-a_2)[(a_1-a_3)(a_1-a_4)(a_1-a_5)-(a_2-a_3)(a_2-a_4)(a_2-a_5)]\ge0

Detailed analysis

For $n=5$ , reorder $a_1\ge a_2\ge a_3\ge a_4\ge a_5$ . The sum of the first two terms is the displayed product. The factor $a_1-a_2$ is nonnegative, and each factor in the first triple product is at least the corresponding factor in the second, so this sum is nonnegative.