Nice. A probabilistic proof. Let $Z$ be a random variable with $\mathbb{P}[Z=i]=a_i$ for $1\le i\le n$. Then $b=\mathbb{E}[Z]$. Let $Z_1,Z_2$ be two iid copies of $Z$. Note that \[ 2{\rm Var}(Z) = \mathbb{E}\bigl[(Z_1-Z_2)^2 \bigr]= \sum_{i\ne j} (i-j)^2 a_i a_j. \]So, it suffices to show \[ {\rm Var}(Z) \le (n-\mathbb{E}[Z])(\mathbb{E}[Z]-1). \]This is nothing but Bhatia-Davis Inequality. For a short proof, simply notice that $(n-Z)(Z-1)\ge 0$, so taking expectations and rearranging, \[ {\rm Var}(Z) \le -n+(n+1)b - b^2 = (n-b)(b-1). \]

It is much simpler: Homogenizing we have the equivalent \[\sum_{i<j} (i-j)^2a_ia_j \le (a_{n-2}+2a_{n-2}+\dots+(n-1)a_1)(a_2+2a_3+3a_4+\dots+(n-1)a_n).\]But $a_ia_j$ appears on the RHS with coefficient $(n-i)(j-1)+(n-j)(i-1)$. So it suffices to check that \[(i-j)^2 \le (n-i)(j-1)+(n-j)(i-1)\]which is equivalent to $(i-1)(n-i)+(j-1)(n-j) \ge 0$ and hence true.

This problem was proposed by Burii.

TheMathBob wrote: Given a positive integer $n\geq 2$ and positive real numbers $a_1, a_2, \ldots, a_n$ with the sum equal to $1$. Let $b = a_1 + 2a_2 + \ldots + n a_n$. Prove that $$\sum_{1\leq i < j \leq n} (i-j)^2 a_i a_j \leq (n-b)(b-1).$$

Attachments:

TheMathBob wrote: Given a positive integer $n\geq 2$ and positive real numbers $a_1, a_2, \ldots, a_n$ with the sum equal to $1$. Let $b = a_1 + 2a_2 + \ldots + n a_n$. Prove that $$\sum_{1\leq i < j \leq n} (i-j)^2 a_i a_j \leq (n-b)(b-1).$$ We have \begin{align*} \mathrm{LHS} &= \frac12\sum_{1\le i,j\le n} (i - j)^2 a_ia_j\\ &= \frac12\sum_{1\le i,j \le n} i^2 a_i a_j - \sum_{1\le i,j\le n} ij a_i a_j + \frac12\sum_{1\le i,j\le n} j^2 a_ia_j\\ &= \sum_{i=1}^n i^2 a_i - b^2. \end{align*}We have $$\mathrm{RHS} - \mathrm{LHS} = (n + 1)b - \sum_{i=1}^n i^2 a_i - n = \sum_{i=1}^n [(n + 1) i - i^2] a_i - n \ge \sum_{i=1}^n n a_i - n = 0$$where we use $(n+1)i - i^2 = n + (n - i)(i - 1) \ge n$.

TheMathBob wrote: Given a positive integer $n\geq 2$ and positive real numbers $a_1, a_2, \ldots, a_n$ with the sum equal to $1$. Let $b = a_1 + 2a_2 + \ldots + n a_n$. Prove that $$\sum_{1\leq i < j \leq n} (i-j)^2 a_i a_j \leq (n-b)(b-1).$$ Very easy problem. $\begin{aligned} \sum_{1\leq i < j \leq n} (i-j)^2 a_i a_j &=\frac 12\sum_{i=1}^n\sum_{j=1}^n(i^2+j^2)a_ia_j-\left(\sum_{k=1}^nka_k\right)^2\\ &=\sum_{i=1}^ni^2a_i\sum_{j=1}^na_j-b^2=\sum_{i=1}^ni^2a_i-b^2\\ &=\sum_{i=1}^n(i-n)(i-1)a_i+\sum_{i=1}^n\left((n+1)i-n\right)a_i-b^2\\ &\leqslant(n+1)\sum_{i=1}^nia_i-n\sum_{i=1}^na_i-b^2\\ &=(n+1)b-n-b^2=(n-b)(b-1).\blacksquare \end{aligned}$