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 1 of 6: Reject every even n

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E_n=\sum_{i=1}^n\prod_{j\ne i}(a_i-a_j),\qquad a_1<0,\ a_2=\cdots=a_n=0\implies E_n=a_1^{n-1}<0

Detailed analysis

Write the expression as $E_n=\sum_{i=1}^n\prod_{j\ne i}(a_i-a_j)$ . For even $n$ , choose $a_1<0$ and $a_2=\cdots=a_n=0$ . Every term except the first contains a repeated zero difference, so $E_n=a_1^{n-1}<0$ because $n-1$ is odd.