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As in the topic, my question is to prove $\sum_{i=1}^{n}\frac{x_{i}^3}{x_{i+1}^2}\geq \sum_{i=1}^{n}\frac{x_{i}^2}{x_{i+1}}$We know that $x_{n+1}=x_{1}$ and $x_{1},x_{2}, x_{3},..., x_{n}\in \mathbb{R}_{+}$.
As I suspect that it would be to cool if all $x_{i}$ were equal, and question is marked as tough one, after a few hours my brain stopped without producing anything reasonable so I ask you for hints how to move it. Everything will be appreciated. Thanks in advance.

3 Answers 3

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Hint:We can derive your inequality by utilizing the Cauchy-Schwarz inequality twice,i.e.,

$(\sum x_i)(\sum\frac{x_i^3}{x_{i+1}^2})\geq (\sum\frac{x_i^2}{x_{i+1}})^2$

and

$(\sum x_i)(\sum\frac{x_i^2}{x_{i+1}})\geq (\sum x_i)^2$

  • 1
    $x_i$ and $x_{i+1}$ are not equal,but their sums are equal,say,$x_1+x_2+\cdots+x_n=x_2+x_3+\cdots+x_{n+1}$.2012-11-11
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$3^n\frac{x_1^3}{x_2^2} + 2\cdot 3^{n-1}\frac{x_2^3}{x_3^2} +4\cdot 3^{n-2}\frac{x_3^3}{x_4^2}+\cdots+2^{n-1}\cdot3\frac{x_{n-1}^3}{x_n^2}+2^n\frac{x_n^3}{x_1^2}\ge (3^n-2^n)x_1$

By Weighted AM-GM. (Note that the sum of the coefficients is the expression for $\frac{3^n-2^n}{3-2} = 3^n-2^n$)

Adding cyclicly and dividing by $3^n-2^n$, we find

$\sum_{i=1}^n\frac{x_i^3}{x_{i-1}^2} \ge \sum_{i=1}^n x_i$

Now note that another application of AM-GM gives $\frac{x_i^3}{x_{i-1}^2}+x_i \ge 2\frac{x_i^2}{x_{i-1}}$, so summing over all $i$,

$\sum_{i=1}^n\frac{x_i^3}{x_{i-1}^2} + \sum_{i=1}^n x_i \ge \sum_{i=1}^n\frac{x_i^2}{x_{i-1}}$

So then, $2\sum_{i=1}^n\frac{x_i^3}{x_{i-1}^2} \ge \sum_{i=1}^n\frac{x_i^3}{x_{i-1}^2} + \sum_{i=1}^n x_i \ge \sum_{i=1}^n\frac{x_i^2}{x_{i-1}}$

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    but this is different from the original inequality2013-11-22
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Multiply both sides by $\prod_{i=1}^{n} x_i^2$ then just use Muirhead inequality.

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    I don't think this works; Muirhead's inequality requires symmetric sums, not just cyclic sums.2012-11-11