Let \color{lime}{n \ge 3} be an integer and let \color{lime}{x_1,x_2,\ldots,x_n} be \color{lime}{n} distinct integers. Prove that
\color{cyan}{(x_1 - x_2)^2 + (x_2 - x_3)^2 + \cdots + (x_n - x_1)^2 \ge 4n - 6}
Using induction:
k=3.
(x_1-x_2)^2+(x_2-x_3)^2+(x_3-x_1)^2\ge 6.
Because the inequality is symmetric, we can assume that x_1<x_2<x_3.
Thus, x_3\ge x_2+1\ge x_1+2.
So, (x_1-x_2)^2\ge 1,(x_2-x_3)^2\ge 1,(x_3-x_1)^2\ge 4.
So, the case for k=3 is true.
Now, assume k=n to be true.
we will have (x_1-x_2)^2+(x_2-x_3)^2+\dots+(x_n-x_1)^2\ge 4n-6. (1)
we need to prove that when k=n+1 the inequality (x_1-x_2)^2+(x_2-x_3)^2+\dots+(x_n-x_{n+1})^2+(x_{n+1}-x_1)^2 \ge 4(n+1)-6 is true as well. (2)
From cycling we can assume that x_{n+1} is the greastes number.
Now, it is enough to prove that (2)-(1) is true.
(2)-(1):
(x_n-x\_{n+1})^2+(x\_{n+1}-x_1)^2-(x_n-x_1)^2\ge 4
2x_{n+1}^2-2x_{n+1}x_n-2x_{n+1}x_1+2x_nx_1\ge4
(x_{n+1}-x_n)(x_{n+1}-x_1)\ge2
And because x_{n+1}> x_n, x_1 and x_n,x_1 are distinct, the inequality is correct and therefore the induction is true.
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