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If $0< a_1\le a_2\le \cdots \le a_n\le a_{n+1}$ and $p>1$, is it true that $$\left(\frac{n+1}{n}\right)^{1-\frac{1}{p}}\left(\frac{\sum_{i=1}^{n+1}a_i^p}{\sum_{i=1}^{n}a_i^p}\right)^{\frac{1}{p}}\ge \frac{\sum_{i=1}^{n+1}a_i}{\sum_{i=1}^{n}a_i}?$$ The numerator and denominator looks like Hölder's inequality.

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Can you tell us why you suspect this? – Anthony Quas Mar 10 '12 at 21:55
If it is true, then I have something to say, but.... – Russel Mar 10 '12 at 22:36
up vote 9 down vote accepted

It's not true. Your proposed inequality can be thought of as saying that the quotient

$(L^p\text{-average of }a_1,\ldots,a_n)/(L^1\text{-average of }a_1,\ldots,a_n)$

is nondecreasing in $n$. If this were true for large $p$ then it would be true for $p=\infty$, which would say that

$a_n/(L^1\text{-average of }a_1,\ldots,a_n)$

is nondecreasing in $n$. But this is clearly false. Just take $a_{n+1}=a_n$: the numerator stays the same but the denominator increases.

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Clever argument. – Russel Mar 10 '12 at 22:36
For example, $a_1 = 1$, $a_2 = a_3 = p = 100$ works. – Woett Mar 11 '12 at 14:20

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