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 3 of 6: Prove the case n=3

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E_3=(a_1-a_2)^2+(a_1-a_3)(a_2-a_3)\ge0\quad(a_1\ge a_2\ge a_3)

Detailed analysis

The expression is symmetric, so reorder $a_1\ge a_2\ge a_3$ . Combining the first two terms gives $(a_1-a_2)^2$ , while the last term is $(a_1-a_3)(a_2-a_3)$ . Both are nonnegative, so $E_3=(a_1-a_2)^2+(a_1-a_3)(a_2-a_3)\ge0$ .