7
$\begingroup$

For $a_1>0$, $a_2>0,\dots,a_n>0$, and $b_1>0$, $b_2>0,\dots,b_n>0$

I want to prove: $$\frac{a_1+a_2+\dots+a_n}{\sqrt{n(b_1+b_2+...+b_n)}} \le \frac{1}{n}\left(\frac{a_1}{\sqrt{b_1}}+\frac{a_2}{\sqrt{b_2}} +\cdots+\frac{a_n}{\sqrt{b_n}}\right)$$

$\endgroup$
  • 3
    $\begingroup$ What do you know? What did you try? $\endgroup$ – Did Jun 25 '11 at 17:37
  • $\begingroup$ @Willie You need something like $a_1 > a_2 > ... > a_n$ and $b_1 < b_2 < ...< b_n$ in this question.. the worst possible rearrangement would be too small otherwise. $\endgroup$ – Zarrax Jun 25 '11 at 18:16
  • $\begingroup$ @Zarrax: thanks, didn't think the problem through and typed on a hunch. $\endgroup$ – Willie Wong Jun 25 '11 at 18:35
5
$\begingroup$

This actually isn't true. Let $r_i = {a_i \over \sum_i a_i}$. Then what you are trying to prove is equivalent to $${1 \over \sqrt{\sum_{i=1}^nb_i \over n}} \leq \sum_{i=1}^n \bigg(r_i {1 \over \sqrt{b_i}}\bigg)$$ Here $\sum_i r_i = 1$ and each $r_i > 0$. But notice that $\sum_{i=1}^n{b_i \over n}$ is the average of the $b_i$. So you are trying to show that $${1 \over \sqrt{b_{average}}} \leq \sum_{i=1}^n \bigg(r_i {1 \over \sqrt{b_i}}\bigg)$$ All you have to do is make the $r_i$ for the largest $b_i$ nearly one, and the others nearly zero, and the inequality won't hold.

So for any choice of $b_i$'s that aren't all the same, you can find some $a_i$'s for which this doesn't work.

$\endgroup$
6
$\begingroup$

For $n=2$ your inequality reduces to

$$\frac{a_1+a_2}{\sqrt{2(b_1+b_2)}} \leq \frac{1}{2} (\frac{a_1}{\sqrt{b_1}}+\frac{a_2}{\sqrt{b_2}})$$

Lets the $a_1=1,a_2=\frac{1}{2},b_1=1,b_2=\frac{11}{16}$ then you have

$$\sqrt{\frac{2}{3}} \leq \frac{1}{2}+\frac{1}{\sqrt{11}}$$

This is false, so the inequality doesn't hold.

$\endgroup$
  • 1
    $\begingroup$ @Aryabhata @Listing $$\sqrt{\frac{2}{3}}=0.81649658092773\ldots$$ $$\frac{1}{2}+\frac{1}{\sqrt{11}}=0.80151134457776\ldots$$ $\endgroup$ – miracle173 Jun 25 '11 at 18:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.