# work 025 – Reflections issue 2010 : S 147

Problem

Let $x_1,x_2,.....,x_n,a,b > 0$ . Prove that the following inequality holds –

$\frac{x_1^3}{(ax_1+bx_2)(ax_2+bx_1)} + \cdots + \frac{x_n^3}{(ax_n+bx_1)(ax_1+bx_n)} \ge \frac{x_1+x_2+x_3+ \cdots x_n}{(a+b)^2}$

My Solution

Set ,

$ax_1 + bx_2 = u_1^2$

$ax_2+bx_1 = v_1^2$

and similarly that for any $k \le n$ ,

$ax_k + bx_{k+1} = u_k^2$

$ax_{k+1}+bx_k = v_k^2$

where, $x_{n+1} = x_1$

we thus have that –

$x_1 + x_2 = \frac{u_1^2+v_1^2}{a+b}$

$x_2+x_3 = \frac{u_2^2+v_2^2}{a+b}$

$\vdots$

$x_n+x_1 = \frac{u_n^2+v_n^2}{a+b}$

Thus,

$2 \sum_{k=1}^n x_k = \frac{1}{a+b}\cdot (\sum_{k=1}^n u_k^2+v_k^2)$

$\implies a+b = \frac{ (\sum_{k=1}^n u_k^2+v_k^2)}{2(\sum_{k=1}^n x_k)}$

$\implies (a+b)^2 = \frac{(\sum_{k=1}^n u_k^2+v_k^2)^2}{4(\sum_{k=1}^n x_k)^2} \ge \frac{ (\sum_{k=1}^n (u_kv_k))^2}{\sum_{k=1}^n x_k}$

Therefore we have ,

$\frac{(\sum_{k=1}^n x_k)^3}{(\sum_{k=1}^n u_kv_k)^2} \ge \frac{x_1+x_2+x_3+ \cdots x_n}{(a+b)^2}$

Thus it is sufficient to prove that –

$\frac{x_1^3}{u_1^2v_1^2} + \cdots + \frac{x_n^3}{u_n^2v_n^2} \ge \frac{(\sum_{k=1}^n x_k)^3}{(\sum_{k=1}^n u_kv_k)^2}$

By Holders’ Inequality ,

$\left(\frac{x_1^3}{u_1^2v_1^2} + \cdots + \frac{x_n^3}{u_n^2v_n^2}\right)( \sum_{k=1}^n u_kv_k)(\sum_{k=1}^n u_kv_k) \ge (\sum_{k=1}^n x_k)^3$

Equality occurs when $x_1=x_2=x_3=\cdots=x_n$

Hence our proof is completed.