Prove inequality: $2\Big[ \sum_{k=0}^{n} (k+1) a_k \Big]^2 -\Big[1+ \sum_{k=0}^{n} a_k \Big]\Big[ \sum_{k=0}^{n} k(k+1)a_k \Big] \geq 0.$

Question

Suppose $a_0\geq\dots \geq a_n \geq 0$ is a sequence of non-negative numbers, where $n+1\leq \sum_{k=0}^{n} a_k \leq n+2$. Then, I want to prove that the following statement is true,

$2\Big[ \sum_{k=0}^{n} (k+1) a_k \Big]^2 -\Big[1+ \sum_{k=0}^{n} a_k \Big]\Big[ \sum_{k=0}^{n} k(k+1)a_k \Big] \geq 0.$

Answer 1

A counterexample is given by the following conditions: $n=185$, $$a_k=\sum_{j=k}^n b_j,\quad b_j:=\frac{c_j}{j+1}, \quad c_j:=\frac{1000}7\,1(j=47)+\frac{302}7\,1(j=185).$$

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Indeed, then $a_0\ge\cdots\ge a_n\ge0$, $$s_0:=\sum_{k=0}^n a_k=\sum_{k=0}^n \sum_{j=k}^n b_j =\sum_{j=0}^n b_j\sum_{k=0}^j 1 \\ =\sum_{j=0}^n b_j(j+1) =\sum_{j=0}^n c_j=\frac{1000}7+\frac{302}7=186\in[n+1,n+2],$$ $$s_1:=\sum_{k=0}^n(k+1)a_k=\sum_{k=0}^n(k+1)\sum_{j=k}^n b_j =\sum_{j=0}^n b_j\sum_{k=0}^j(k+1) \\ =\sum_{j=0}^n b_j\frac{(j+1)(j+2)}2 =\sum_{j=0}^n c_j\frac{j+2}2=\frac{105474}7,$$ $$s_2:=\sum_{k=0}^n k(k+1)a_k=\sum_{k=0}^n k(k+1)\sum_{j=k}^n b_j =\sum_{j=0}^n b_j\sum_{k=0}^j k(k+1) \\ =\sum_{j=0}^n b_j\frac{j(j+1)(j+2)}3 =\sum_{j=0}^n c_j\frac{j(j+2)}3=\frac{4250230}7,$$ so that the left-hand side of your inequality is $$2s_1^2-(1+s_0)s_2=-\frac{1168732}{49}<0.\quad\Box$$