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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 .

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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. – Simon Hayward Nov 29 '12 at 20:28
At least post what the question is. It doesn't have to be perfect, some else can edit it. – Simon Hayward Nov 29 '12 at 20:36
Right. I didn't -1 it. But I will now. – Simon Hayward Nov 29 '12 at 20:38
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. – Did Nov 29 '12 at 21:41
The original question consisted of only a ink to that question. – Simon Hayward Nov 30 '12 at 9:01
up vote 6 down vote accepted

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.

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