Show that $\frac {a_1^2}{a_2}+\frac {a_2^2}{a_3}+...+\frac {a_n^2}{a_1}\geq a_1+a_2+...+a_n$ using AM-GM.

Question

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.

score 24 · Accepted Answer · edited Jul 10 '13 at 15:47

24

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

add all inequalities and you're done!

edited Jul 10 '13 at 15:47

dajoker

2,297
14
25

answered May 16 '13 at 12:56

math110

93,304

and this equality can use by Cauchy-Schwarz inequality.and have other nice methods,Thank you – math110 May 16 '13 at 12:56
1

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

score 5 · Answer 2 · answered May 16 '13 at 13:09

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

Show that $\frac {a_1^2}{a_2}+\frac {a_2^2}{a_3}+...+\frac {a_n^2}{a_1}\geq a_1+a_2+...+a_n$ using AM-GM.

2 Answers2