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
dnkywin
16.04.2012 20:45
Claim: the maximum value is given by the function $f(m,n,r,s) = \min(m,n)^{\frac1r - \frac 1s}$, where equality holds when $a_{ij}=1$ if $i=j$ and $a_{ij}=0$ otherwise.
First off, let $b_{ij} = a_{ij}^r$, $k = s/r$. It suffices to show that
\[\sum_{j=1}^n\sqrt[k]{\sum_{i=1}^m b_{ij}^k} \le \min(m,n)^{1-\frac1k}\left(\sqrt[k]{\sum_{i=1}^m \left(\sum_{j=1}^n b_{ij}\right)^k}\right) \]
Now we need the following lemma:
Lemma: If $c_1\ge c_2\ge\ldots\ge c_n\ge0$ and $x_1,x_2,\ldots,x_n\ge0$, $k\ge1$, and if $S = \sum_{i=1}^n x_i$, then
\[\sum_{i=1}^n \sqrt[k]{c_i+x_i^k} \le \sqrt[k]{c_n+S^k} + \sum_{i=1}^{n-1}\sqrt[k]{c_i} \]
Proof: If $c_n=0$ then the above inequality is obvious, as $\sqrt[k]{c_i+x_i^k}\le x_i + \sqrt[k]{c_i}$ for all $i$. Now if $c_n>0$, notice that the above inequality is equivalent to
\[\sum_{i=1}^n \lambda_i\sqrt[k]{1 + \left(\frac{x_i}{\lambda_i}\right)^k} \le \lambda_n\sqrt[k]{1 + \left(\frac{S}{\lambda_n}\right)^k} + \sum_{i=1}^{n-1} \lambda_i \]
where $\lambda_i = \sqrt[k]{c_i}$. However, if $g(x) = \sqrt[k]{x^k+1}$, $g$ is convex on the positive reals, as $g''(x) = (k-1)x^{k-2}(x^k+1)^{\frac1k-2}\ge0$ for all positive real numbers $x$. In addition, $\frac{S}{\lambda_n}\ge \frac{x_i}{\lambda_i}$ for any $i$, so the above inequality is indeed true by Karamata's inequality. $\boxed{}$
Let $S_i=\sum_{j=1}^n b_{ij}$. By our lemma, in the left hand side of our original inequality, we can 'smooth' all of the $b_{ij}$'s to $b'_{ij}$ such that for each $i$ there is a unique $j$ such that $b'_{ij} = S_i$ and $b_{ij} = 0$ for all other $j$ without decreasing the LHS of the original inequality, and so there exists a partition $P = \{A_1,\ldots,A_l\}$ of $\{{1,2,\ldots,m\}}$ such that
\[\sum_{j=1}^n\sqrt[k]{\sum_{i=1}^m b_{ij}^k} \le\sum_{j=1}^n\sqrt[k]{\sum_{i=1}^m b'_{ij}^k} = \sum_{j=1}^l \sqrt[k]{\sum_{i\in A_j} S_i^k}\]
However, by Power Mean,
\[\sum_{j=1}^l \sqrt[k]{\sum_{i\in A_j} S_i^k}\le l^{1-\frac1k}\sqrt[k]{\sum_{j=1}^l\left(\sqrt[k]{\sum_{i\in A_j} S_i}\right)^k}=l^{1-\frac1k}\left(\sqrt[k]{\sum_{i=1}^m S_i^k}\right) \]
But $l\le n$ (as we never increase the number of vectors) and $l\le m$ (as each new vector corresponds to at least one unique dimension), so the original inequality follows.
soshi
03.03.2013 19:48
are you sure that's correct, you didn't consider the tensor products
P-H-David-Clarence
07.11.2016 02:56
Is that correct?
l1090107005
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.
EthanWYX2009
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.