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.

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