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.



A Hölder like inequality
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.


