Another mixed mean inequality

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Let $a_1,$ $a_2,$ $\ldots,$ $a_n$ be positive real numbers. Prove that $$\sqrt{\frac{a_1^2+\left( \frac{a_1+a_2}{2}\right)^2+\cdots +\left(\frac{a_1+a_2+\cdots +a_n}{n}\right)^2}{n}} \le \frac{a_1+\sqrt{\frac{a_1^2+a_2^2}{2}}+\cdots+\sqrt{\frac{a_1^2+a_2^2+\cdots +a_n^2}{n}}}{n}.$$

I have proved this inequality for $n=2$ and $n=3.$ But I still cannot prove it for the general case. Can somone help me?

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I came across a monthly article by a graduate student, which has a strong resemblance to the one you ask. but I cannot remember which one. – Sunni May 19 2010 at 16:58

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You'll probably have more luck at the art of problem solving forums artofproblemsolving.com/Forum/index.php – jc May 19 2010 at 18:42

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I like the question and am tempted to have a go -- but no time right now. – gowers May 19 2010 at 22:54

1 Answer

Mixed mean inequalities have been studied quite a bit, inspired mostly by the inequalities of Carleman and Hardy, starting probably from this article of K. Kedlaya. Note that the following holds if $r < s$ (in your case $r=1, s=2$): $$\left(\frac{1}{n}\sum_{k=1}^n \left(M_k^{[r]}(\mathbf a)\right) ^s\right)^{1/s}\leq \left(\frac{1}{n}\sum_{k=1}^n\left(M_k^{[s]}(\mathbf a)\right) ^r \right)^{1/r}$$ Where $\mathbf a$ denotes a sequence of real numbers, and $M_k^{[r]}$ denotes the $r$-th power mean of the first $k$ variables. You can find a proof of the general form in "Survey on classical inequalities" by T.M. Rassias (page 32, it is the original proof of B. Mond and J. Pecaric which was later extended to matrices and linear operators).

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Thanks Gjergji Zaimi for your reference. – can_hang2007 Jun 2 2010 at 1:38
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