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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.

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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

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

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  • $\begingroup$ Can you explain second inequality, please? $\endgroup$ – fdhd Nov 11 '12 at 14:10
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    $\begingroup$ @user46034:It is just $$(\sum x_{i+1})(\sum\frac{x_i^2}{x_{i+1}})\geq (\sum x_i)^2$$ then it is easy to use Cauchy-Schwartz inequality to get the result above. $\endgroup$ – zy_ Nov 11 '12 at 14:14
  • $\begingroup$ Why do you change $x_{i}$ with $x_{i+1}$? Are they equal? Or Cauchy-...? I really do not understand $\endgroup$ – fdhd Nov 11 '12 at 14:17
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    $\begingroup$ $x_i$ and $x_{i+1}$ are not equal,but their sums are equal,say,$x_1+x_2+\cdots+x_n=x_2+x_3+\cdots+x_{n+1}$. $\endgroup$ – zy_ Nov 11 '12 at 14:19
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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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  • $\begingroup$ but this is different from the original inequality $\endgroup$ – Christmas Bunny Nov 22 '13 at 0:51
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Multiply both sides by $$\prod_{i=1}^{n} x_i^2$$ then just use Muirhead inequality.

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  • $\begingroup$ I don't think this works; Muirhead's inequality requires symmetric sums, not just cyclic sums. $\endgroup$ – only Nov 11 '12 at 13:50

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