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Sum of squares >= than sum of cubes

Source: Romanian NO, grade ix, p.1

October 3, 2019
inequalitiesformulasalgebrafactorizationsum of cubes

Problem Statement

Let be a natural number n n and n n real numbers a1,a2,…,an a_1,a_2,\ldots ,a_n such that a_m+a_{m+1} +\cdots +a_n\ge \frac{(m+n)(n-m+1)}{2} , \forall m\in\{ 1,2,\ldots ,n \} . Prove that a12+a22+⋯+an2≥n(n+1)(2n+1)6. a_1^2+a_2^2+\cdots +a_n^2\ge\frac{n(n+1)(2n+1)}{6} .
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