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Algebra Difficulty 4.8 AIME Prove it Ukraine

Prove that for natural numbers a≥b≥c≥da \ge b \ge c \ge d the inequality:
ab+bc+cd−b2−c2−d2≥a−d. ab+bc+cd-b^{2}-c^{2}-d^{2} \ge a-d.

Solution

We have:
ab+bc+cd−b2−c2−d2=b(a−b)+c(b−c)−d(c−d)≥≥(a−b)+(b−c)+(c−d)=a−d. ab+bc+cd-b^{2}-c^{2}-d^{2} = b(a-b)+c(b-c)-d(c-d) \ge \\ \ge (a-b)+(b-c)+(c-d)=a-d.

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