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 6 of 6: Group the last two terms and conclude

Previous step Next step

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

Detailed analysis

The last two terms equal the displayed expression. Here $a_4-a_5\ge0$ , and each factor in the first triple product is at least its counterpart in the second, so this sum is nonnegative. Together with the preceding groups, $E_5\ge0$ . Thus the assertion holds exactly for $n=3$ and $n=5$ .