Given two integers $m,n$ which are greater than $1$. $r,s$ are two given positive real numbers such that $r<s$. For all $a_{ij}\ge 0$ which are not all zeroes,find the maximal value of the expression \[f=\frac{(\sum_{j=1}^{n}(\sum_{i=1}^{m}a_{ij}^s)^{\frac{r}{s}})^{\frac{1}{r}}}{(\sum_{i=1}^{m})\sum_{j=1}^{n}a_{ij}^r)^{\frac{s}{r}})^{\frac{1}{s}}}.\]
Problem
Source: 2012 China TST Test 2 p5
Tags: function, inequalities, vector, inequalities proposed
16.04.2012 20:45
03.03.2013 19:48
are you sure that's correct, you didn't consider the tensor products
07.11.2016 02:56
Is that correct?
04.07.2018 10:14
Sorry to bumping this up. But I'm wondering whether it is true: For $x_1,x_2,\cdots,x_n>0,r<s$,we have:$$\left(\sum_{i=1}^n x_i^s\right)^{\frac{1}{s}}\le\left(\sum_{i=1}^n x_i^r\right)^{\frac{1}{r}}$$ If so, the problem can be killed immediately.
17.05.2023 10:42
See also here We prove that the answer is $\boxed{\min\{m,n\}^{\frac 1r-\frac1s}}$. For $1\leq k\leq\min{m,n},$ let $a_{kk}=1$ and otherwise let it be ${}{0}$. Then $f=\min\{m,n\}^{\frac 1r-\frac1s}$. For $1\leq j\leq n$, let $x_j=\left( \sum\limits_{i=1}^ma_{ij}^s \right) ^{\frac{1}{s}}$. For $1\leq i\leq m$, let $y_i=\left( \sum\limits_{j=1}^na_{ij}^r \right) ^{\frac{1}{r}}$. Then $$f=\dfrac{\left( \sum\limits_{j=1}^nx_j^r \right) ^{\frac{1}{r}}}{\left( \sum\limits_{i=1}^my_i^s\right) ^{\frac{1}{s}}}.$$First we have $x_j\leqslant\left( \sum\limits_{i=1}^ma_{ij}^r \right) ^{\frac{1}{r}}\Rightarrow\sum\limits_{j=1}^nx_j^r\leqslant\sum\limits_{j=1}^n\sum\limits_{i=1}^ma_{ij}^r=\sum\limits_{i=1}^m\sum\limits_{j=1}^na_{ij}^r=\sum\limits_{i=1}^my_i^r$. Then $$f=\dfrac{\left( \sum\limits_{j=1}^nx_j^r \right) ^{\frac{1}{r}}}{\left( \sum\limits_{i=1}^my_i^s\right) ^{\frac{1}{s}}}\leqslant\dfrac{\left( \sum\limits_{i=1}^my_i^r \right) ^{\frac{1}{r}}}{\left( \sum\limits_{i=1}^my_i^s\right) ^{\frac{1}{s}}}\leqslant m^{\frac1r-\frac1s}.$$Second we have $y_i\geqslant\left( \sum\limits_{j=1}^na_{ij}^s \right) ^{\frac{1}{s}}\Rightarrow\sum\limits_{i=1}^my_i^s\geqslant\sum\limits_{i=1}^m\sum\limits_{j=1}^na_{ij}^s=\sum\limits_{j=1}^n\sum\limits_{i=1}^ma_{ij}^s=\sum\limits_{j=1}^nx_j^s$. Then $$f=\dfrac{\left( \sum\limits_{j=1}^nx_j^r \right) ^{\frac{1}{r}}}{\left( \sum\limits_{i=1}^my_i^s\right) ^{\frac{1}{s}}}\leqslant\dfrac{\left( \sum\limits_{j=1}^nx_j^r \right) ^{\frac{1}{r}}}{\left( \sum\limits_{j=1}^nx_j^s\right) ^{\frac{1}{s}}}\leqslant n^{\frac1r-\frac1s}.$$Therefore $f\leqslant\min\{m,n\}^{\frac 1r-\frac1s}$. Typing $\LaTeX$ is too tiring.