3

Let $x_{i}>0(i=1,2,\cdots,n)$,and such $x_{1}+x_{2}+\cdots+x_{n}=1$,show that $$n-1+\sum_{i=1}^{n}\dfrac{x^2_{i}}{x_{i+1}}\ge\sum_{i=1}^{n}\dfrac{1}{x_{i+1}-x_{i}+1}\tag{1}$$ where $x_{n+1}=x_{1}$

I have use Cauchy-schwarz inequality $$LHS\ge n-1+\dfrac{(x_{1}+x_{2}+\cdots+x_{n})^2}{x_{1}+x_{2}+\cdots+x_{n}}=n$$ so I think maybe have $$\sum_{i=1}^{n}\dfrac{1}{x_{i+1}-x_{i}+1}\le n?$$ But it's a pity that I finally found out use C-S $$\sum_{i=1}^{n}\dfrac{1}{x_{i+1}-x_{i}+1}\ge\dfrac{(1+1+\cdots+1)^2}{\sum_{i=1}^{n}(x_{i+1}-x_{i}+1)}=n$$

So far, I still can't solve this problem (1).

math110
  • 93,304

1 Answers1

2

First off, subtract $n$ from both sides to get:

$$LHS = \sum_{i=1}^n\dfrac{x_i(x_i-x_{i+1})}{x_{i+1}}\geq\sum_{i=1}^n\dfrac{x_i-x_{i+1}}{x_{i+1}-x_i+1} = RHS.$$ Now subtract RHS from LHS:

$$LHS-RHS = \sum_{i=1}^n(x_i-x_{i+1})\left(\dfrac{x_i}{x_{i+1}} - \dfrac{1}{x_{i+1}-x_i+1}\right) = \sum_{i=1}^n\dfrac{(x_i-x_{i+1})^2(1-x_i)}{x_i(x_{i+1}-x_i+1)}\geq 0.$$

dezdichado
  • 13,888