# Problem with inequality $\min (x_1,x_2,\ldots,x_n)$

let $0\le x_i$, $i=1,2,\ldots,n$, and $a_i=1+(i-1)d$, $d\in[0,2],\forall i\in\{1,2,3,\ldots,n\}$, show that $$(1+a_n)\left(x_1+x_2+\cdots+x_n\right)^2\ge 2n \min(x_1,x_2,\ldots,x_n) \left(\sum_{i=1}^n a_i x_i\right)$$

$n=1,2,3$ is not hard to prove it,I am unsure what to do from here, I know I somehow need to compare a form of this expression to $\min{(x_1,x_2,\ldots,x_n)}$?, but how? Am I on the right lines?but I don't have any idea how to start proving it

• Have you tried induction? – Sergio Parreiras Aug 3 '15 at 14:53
• $(x_1+\cdots+x_n) \geq n \cdot \min(x_1.\cdots+x_n)$.. for $d=2$, $1+a_n=2n$ and for $d=0$ all $a_i=1$ and $1+a_n=2$. Also $a_n\geq a_i$ for all $i$. – vudu vucu Aug 3 '15 at 15:02

First, by the rearrangement inequality, the sum on the right side is maximized when $x_i$'s are arranged in ascending order. Thus, it suffices to show the case with $x_1 \le x_2 \le \cdots \le x_n$. In other words, we wish to show $$(1+a_n)(x_1 + \cdots + x_n)^2 \ge 2 \, n \, x_1 \, (a_1 \, x_1 + \cdots + a_n \, x_n).$$
Second, suppose $x_2 > x_1$, we can replace $x_n$ by $x_n + x_2 - x_1$, and then replace $x_2$ by $x_1$. This would leave sum $x_1 + \cdots + x_n$ unchanged, while increasing the sum on the right-hand side. We can also do this for $x_3, \dots, x_{n-1}$. This means we only need to show the case with $x_1 = x_2 = \cdots = x_{n-1} \le x_n$. Define $\Delta \equiv x_n - x_1$, we only need to show that \begin{align} (1+a_n)(n \, x_1 + \Delta)^2 \ge 2 \, n \, x_1 \, (a_1 \, x_1 + \cdots + a_n \, x_1 + a_n \, \Delta) \\ = 2 \, n \, x_1 \, \left[ \frac{1}{2}(1 + a_n) \, n \, x_1 + a_n \, \Delta\right]. \end{align}