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 2 of 6: Reject odd n at least 7

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c>a>b,\quad a_1=a,\ a_2=a_3=a_4=b,\ a_5=\cdots=a_n=c\implies E_n=(a-b)^3(a-c)^{n-4}<0

Detailed analysis

For odd $n\ge7$ , choose $c>a>b$ , set $a_1=a$ , $a_2=a_3=a_4=b$ , and $a_5=\cdots=a_n=c$ . Repeated values make all terms except the first vanish, giving $E_n=(a-b)^3(a-c)^{n-4}$ . Since $n-4$ is odd, this is negative.