3
$\begingroup$

Assume that $a_1, \dots,a_n $ and $b_1, \dots,b_n$ are $2n$ non-negative real numbers.

We have $$\sum_{i=1}^na_i = \sum_{i=1}^nb_i$$

We're to prove that $$\sqrt2 \sum_{i=1}^n (\sqrt{a_i}-\sqrt {b_i})^2 \ge \sum_{i=1}^n|a_i-b_i|.$$ Can anyone help!

I encountered it while i was surfing in olympiad section of artofproblemsolving and found it interesting , since my olympiads are very near so I tried to solve this inequality but failed to do so. I tried to apply AM-GM-HM Inequality but it doesnt works here & also tried Cauchy-Schwarz & Tchebycheff's Inequality too but with no success . I just cant figure out what to keep as variables in the formulae stated above .

$\endgroup$
11
  • $\begingroup$ Are you kidding me? You are posting a link to your actual question? Please improve this or it's going to end up being deleted as a low quality post. $\endgroup$ Nov 29, 2012 at 20:28
  • 3
    $\begingroup$ At least post what the question is. It doesn't have to be perfect, some else can edit it. $\endgroup$ Nov 29, 2012 at 20:36
  • 1
    $\begingroup$ Right. I didn't -1 it. But I will now. $\endgroup$ Nov 29, 2012 at 20:38
  • 2
    $\begingroup$ Now that the question is readable... Welcome to math.SE: since you are new, you might want to know that, in order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Please consider rewriting your post. $\endgroup$
    – Did
    Nov 29, 2012 at 21:41
  • 1
    $\begingroup$ The original question consisted of only a ink to that question. $\endgroup$ Nov 30, 2012 at 9:01

1 Answer 1

7
$\begingroup$

If $n=2$ and $a_1=b_2=100, a_2=b_1=121$, then the inequality becomes $2\sqrt{2}\ge 42$, which is false. So the inequality does not actually hold.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .