Inequality with interesting independent constants Let $b_1,\dots,b_{n-1}$ be integers satisfying $0 \le b_i \le n-i$ for each $i \in [n-1]$ such that $\sum_{i=1}^{n-1} b_i = \alpha \binom{n}{2}$ where $\alpha$ is constant strictly between $0$ and $1$. Prove that $\sum_{i=1}^{n-1} ib_i \ge c\alpha^2(n-1)\binom{n}{2}$ for some constant $c$ which is independent of $\alpha$ and $n$.
My approach is as follows: I basically tried to split the entire sum into $n-1$ terms like $b_k+\dots+b_{n-1}$ and bound it by using $b_i \le n-i$ as follows $b_k+\dots+b_{n-1}=b_1+\dots+b_{n-1}-(b_1+\dots+b_{k-1})\ge \alpha\binom{n}{2}-(n-1+\dots+n-k+1)=\alpha\binom{n}{2}-\frac{(k-1)(2n-k)}{2}$ (for $k \ge 2$). However, I couldn't finish this idea. Any help is appreciated! Thank you!
 A: Trivially the constant $c=0$ will satisfy the inequality for any $\alpha$ and $n$. For the following, I will pick $c = \frac 14$.
For the proof, I will relax the requirement of $\alpha$ to $0\le \alpha \le 1$. Hence every $b_1, \ldots, b_{n-1}$ may satisfy $0\le b_i \le n-i$ without further global constraints.
I will also rewrite the RHS of the inequality to be proven to
$$c\alpha^2 (n-1)\binom n2 = \frac{2c}{n}\alpha^2 \frac{n(n-1)}2\binom n2  = \frac{2c}{n}\left(\sum_{i=1}^{n-1} b_i\right)^2$$

For $n=1$, $0\ge 0$ is true.
For $n=2$, $0\le b_1 \le 1$ (which would not satisfy the stricter $\alpha$ requirement in the question), so
$$RHS = cb_1^2 = \frac{b_1}{4}b_1 \le b_1 = LHS$$
Induction hypothesis: assume that for some $n=k$, if there are $0\le b_i\le k-i$ for every $i=1,\ldots, k-1$, then
$$\sum_{i=1}^{k-1}ib_i \ge \frac{2c}{k}\left(\sum_{i=1}^{k-1}b_i\right)^2$$

For $n=k+1$, the $b_i$s satisfy $0\le b_i \le k+1-i$ for every $i=1,\ldots, k$. This means that, omitting $b_1$, the remaining $b_i$s satisfy $0\le b_{j+1}\le k-j$ for every $j = 1,\ldots, k$.
Let the $S$ be the sum of the remaining $b_i$s: $S = \sum_{i=2}^k b_i = b_2 + b_3 + \cdots + b_k$.
$$\begin{align*}
LHS &= \sum_{i=1}^k ib_i\\
&= \sum_{i=1}^k b_i + \sum_{i=2}^k (i-1)b_i\\
&= b_1 + S + \sum_{j=1}^{k-1} jb_{j+1}&&j=i-1\\
&\ge b_1 + S + \frac{2c}{k}\left(\sum_{j=1}^{k-1}b_{j+1}\right)^2 &&\text{induction hypothesis}\\
&= b_1 + S + \frac{2c}k S^2\\
RHS &= \frac{2c}{k+1} \left(\sum_{i=1}^kb_i\right)^2
= \frac{2c}{k+1} \left(b_1+S\right)^2
\end{align*}$$
Comparing the difference between both sides,
$$\begin{align*}
LHS - RHS &\ge \left(b_1+S+\frac{2c}k S^2\right) - \frac{2c}{k+1} \left(b_1+S\right)^2\\
&= b_1+S+\frac{2c}k S^2 - \frac{2c}{k+1}b_1^2 - \frac{4c}{k+1}b_1S - \frac{2c}{k+1}S^2\\
&= \left(1-2c\frac{b_1}{k+1}\right)b_1 + \left(1-4c\frac{b_1}{k+1}\right)S  + 2c\left(\frac1k - \frac1{k+1}\right)S^2\\
&\ge 0\\
LHS &\ge RHS
\end{align*}$$
using $0\le b_1 \le k$, $0\le S$, and $0\le c \le \frac14$.

By induction, the inequality holds for all positive integer $n$ with appropriate $b_i$s, using constant $c=\frac 14$. The same inequality would also hold with additional constraint on $\alpha$.
