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Let $a_1,...,a_n$ be given positive reals, such that:

$$\sum a_i \times \sum \frac1{a_i} \le (n + \frac12)^2$$

Prove that $\max \{a_i\} \le 4 \min \{a_i\}$

I don't know exactly how to approach this. I wrote it equivalently:

$$\frac{\operatorname{AM}}{\operatorname{HM}} n^2 \le (n + \frac12)^2$$

And expanded from there to reach nothing

Thank you.

  • Dear downvoter, please enlighten me; I'm not wise enough to know why did you take such a decision. –  Aug 25 '15 at 07:17

1 Answers1

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Let $a_1 \ge a_2 \ge ...\ge a_n$ then using CS

$$(n+\dfrac 1 2)^2\ge (\sum_{i=1}^n a_i)(\sum_{i=1}^n \dfrac 1 {a_i})\ge \\ \ge (\sqrt{\dfrac {a_n} {a_1}}+\sqrt{\dfrac {a_1} {a_n}}+n-2)^2 \Rightarrow$$

$$\sqrt{\dfrac {a_n} {a_1}}+\sqrt{\dfrac {a_1} {a_n}}\le \dfrac 5 2$$

Let $\sqrt{\dfrac {a_1} {a_n}}=x$ then

$$x+\dfrac 1 x\le \dfrac 5 2 \Rightarrow (x-\dfrac 1 2)(x-2)\le 0 \Rightarrow$$

$$x\le 2\Rightarrow a_1 \le 4a_n$$

Booldy
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  • I thought about CS but I looked at it as: $\sum a_i \sum 1/a_i = \sum (\sqrt{a_i})^2 \sum (\sqrt{1/a_i})^2 \ge (\sum 1)^2 = n^2$. How did you use CS for your result? –  Aug 25 '15 at 09:28
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    $(a_1+a_2+...+a_n)\cdot (\dfrac 1 {a_n}+\dfrac 1 {a_2}+...+\dfrac 1 {a_1})$ – Booldy Aug 25 '15 at 09:42
  • As far as I can get, $(a_1 + ... + a_n)(1/a_n + ... + 1/a_1) \ge \sqrt{\frac{a_1}{a_n}} + \sqrt{\frac{a_n}{a_1}} + \sum_{i \neq 1,n} \sqrt{\frac{a_i}{a_{n-i+1}}}$. But why is that sum $\ge n -2$? I understand that $a_1 \le ... \le a_n$, but this can be only useful up to $(n-1)/2$ or $n/2$. What is the trick that you used? –  Aug 25 '15 at 09:57
  • CS inequality $(x_1^2+x_2^2+...+x_n^2)(y_1^2+y_2^2+...+y_n^2)\ge(x_1 y_1+...+x_ny_n)^2$ now we chose $x_1=\sqrt{a_1},x_2=\sqrt{a_2},...,x_n=\sqrt{a_n},y_1=\sqrt{\dfrac 1 {a_n}},y_2=\sqrt{\dfrac 1 {a_2}},y_3=\sqrt{\dfrac 1 {a_3}},...,y_n=\sqrt{\dfrac 1 {a_1}}$ and just apply it – Booldy Aug 25 '15 at 10:05
  • alright, now it's clear. I thought that you reversed the whole sum (making it $1/a_n + $ $1/a_{n-1}$ $ + ... + 1/a_1$. That was very nice by the way! –  Aug 25 '15 at 10:08