Given $a_1,a_2,...,a_n$ be positive reals. Show that $\displaystyle\frac {a_1^2}{a_2}+\frac {a_2^2}{a_3}+...+\frac {a_n^2}{a_1}\geq a_1+a_2+...+a_n$ using AM-GM.

I know how to slve it using rearrangement inequality, but I can't. How should I apply AM-GM? Thanks.


closed as off-topic by Carl Mummert, Brahadeesh, user21820, Namaste, Did Dec 11 '18 at 19:22

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$$\dfrac{a^2_{1}}{a_{2}}+a_{2}\ge 2a_{1}$$ $$\dfrac{a^2_{2}}{a_{3}}+a_{3}\ge 2a_{2}$$ $$\cdots\cdots$$ $$\dfrac{a^2_{n}}{a_{1}}+a_{1}\ge 2a_{n}$$

add all inequalities and you're done!

  • $\begingroup$ and this equality can use by Cauchy-Schwarz inequality.and have other nice methods,Thank you $\endgroup$ – math110 May 16 '13 at 12:56
  • 1
    $\begingroup$ use this cauchy-Schwarz $$\left(\dfrac{a^2_{1}}{a_{2}}+\dfrac{a^2_{2}}{a_{3}}+\cdots+\dfrac{a^2_{n}}{a_{1}}\right)(a_{2}+a_{3}+\cdots+a_{n}+a_{1})\ge (a_{1}+a_{2}+\cdots+a_{n})^2$$ $\endgroup$ – math110 May 16 '13 at 13:00
  • $\begingroup$ Nice. What do you mean by "and all equality"? $\endgroup$ – Julien May 16 '13 at 13:32
  • $\begingroup$ Thank you, I mean add,@julien $\endgroup$ – math110 May 16 '13 at 13:45

other nice methods $$\Longleftrightarrow \left(\dfrac{a^2_{1}}{a_{2}}-2a_{1}+a_{2}\right)+\left(\dfrac{a^2_{2}}{a_{3}}-2a_{2}+a_{3}\right)+\cdots+\left(\dfrac{a^2_{n}}{a_{1}}-2a_{n}+a_{1}\right)\ge 0$$ $$\Longleftrightarrow \left(\dfrac{a_{1}}{\sqrt{a_{2}}}-\sqrt{a_{2}}\right)^2+\cdots+\left(\dfrac{a_{n}}{\sqrt{a_{1}}}-\sqrt{a_{1}}\right)^2\ge 0$$


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