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How to prove this inequality?
$$\left(\frac{x_1}{x_2}\right)^{4}+\left(\frac{x_2}{x_3}\right)^{4}+\dotsb+\left(\frac{x_n}{x_1}\right)^{4}\geq \frac{x_1}{x_5}+\frac{x_2}{x_6}+\dotsb+\frac{x_{n-3}}{x_1}+\frac{x_{n-2}}{x_2}+\frac{x_{n-1}}{x_3}+\frac{x_n}{x_4},$$ where $x_1, x_2, \dotsc, x_n>0.$

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  • $\begingroup$ What is the source of this problem? $\endgroup$ – wythagoras Apr 21 '17 at 15:24
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    $\begingroup$ sounds somewhat like an AMGM generalization to me... try doing some induction too $\endgroup$ – jorgeegroj Apr 21 '17 at 15:26
  • $\begingroup$ how does it look if $$n=2,3$$? $\endgroup$ – Dr. Sonnhard Graubner Apr 21 '17 at 15:43
  • $\begingroup$ @Dr. Sonnhard Graubner, only for n=5, 6, ... $\endgroup$ – Doublench Apr 21 '17 at 15:46
  • $\begingroup$ how it looks then? $\endgroup$ – Dr. Sonnhard Graubner Apr 21 '17 at 15:46
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Let $\frac{x_1}{x_2}=y_1$, $\frac{x_2}{x_3}=y_2$, $\frac{x_3}{x_4}=y_3$, $\frac{x_4}{x_5}=y_4$,...

Hence, $\prod\limits_{i=1}^ny_i=\frac{x_1}{x_{n+1}}$ and we need to prove that $$\sum_{i=1}^ny_i^4\geq\sum_{i=1}^ny_iy_{i+1}y_{i+2}y_{i+3},$$ which is just AM-GM: $$\sum_{i=1}^ny_i^4=\frac{1}{4}\sum_{i=1}^n(y_i^4+y_{i+1}^4+y_{i+2}^4+y_{i+3}^4)\geq\sum_{i=1}^ny_iy_{i+1}y_{i+2}y_{i+3}$$. Done!

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  • $\begingroup$ Lol. Genius. Thx! $\endgroup$ – Doublench Apr 21 '17 at 16:22
  • $\begingroup$ How $$\sum_{i=1}^ny_i^4=\frac{1}{4}\sum_{i=1}^n(y_i^4+y_{i+1}^4+y_{i+_2}^4+y_{i+3}^4)$$? $\endgroup$ – Doublench Apr 21 '17 at 16:46
  • $\begingroup$ @Doublench Because $\sum\limits_{i=1}^ny_i^4=\sum\limits_{i=1}^ny_{i+1}^4=\sum\limits_{i=1}^ny_{i+2}^4=\sum\limits_{i=1}^ny_{i+3}^4$ $\endgroup$ – Michael Rozenberg Apr 21 '17 at 16:54
  • $\begingroup$ nope man. How ? For example, n=5. $\endgroup$ – Doublench Apr 21 '17 at 17:01
  • $\begingroup$ @Doublench For $n=5$ we have $y_1^4+y_2^4+y_3^4+y_4^4+y_5^4=\frac{1}{4}((y_1^4+y_2^4+y_3^4+y_4^4)+(y_2^4+y_3^4+y_4^4+y_5^4)+(y_3^4+y_4^4+y_5^4+y_1^4)+(y_4^4+y_5^4+y_1^4+y_2^4)+(y_5^4+y_1^4+y_2^4+y_3^4))$ $\endgroup$ – Michael Rozenberg Apr 21 '17 at 17:06

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