Fang-jh wrote: For two given positive integers $ m,n > 1$, let $ a_{ij} (i = 1,2,\cdots,n,j = 1,2,\cdots,m)$ be nonnegative real numbers, not all zero, find the maximum and the minimum values of $ f$, where $ f = \frac {n\sum_{i = 1}^{n}(\sum_{j = 1}^{m}a_{ij})^2 + m\sum_{j = 1}^{m}(\sum_{i = 1}^{n}a_{ij})^2}{(\sum_{i = 1}^{n}\sum_{j = 1}^{m}a_{ij})^2 + mn\sum_{i = 1}^{n}\sum_{i = j}^{m}a_{ij}^2}$. Dear fang-jh, I was wondering if there was a typo in the denominator in this term: $ \sum_{i = 1}^{n}\sum_{i = j}^{m}a_{ij}^2$ .. i guess its a valid sum, but then the (inner) sum should be written from $ j=i$ to $ j=m$, not from $ i = j$ to $ i=m$ ; and the (outer) sum from $ i = 1$ to $ m$, not $ i = 1$ to $ n$ (since if $ n \geq i \geq m$, the second sum is empty), and it is quite a strange sum. Perhaps it may be this: $ \sum_{i = 1}^{n}\sum_{j = 1}^{m}a_{ij}^2$?

epitomy01 wrote: Fang-jh wrote: For two given positive integers $ m,n > 1$, let $ a_{ij} (i = 1,2,\cdots,n,j = 1,2,\cdots,m)$ be nonnegative real numbers, not all zero, find the maximum and the minimum values of $ f$, where $ f = \frac {n\sum_{i = 1}^{n}(\sum_{j = 1}^{m}a_{ij})^2 + m\sum_{j = 1}^{m}(\sum_{i = 1}^{n}a_{ij})^2}{(\sum_{i = 1}^{n}\sum_{j = 1}^{m}a_{ij})^2 + mn\sum_{i = 1}^{n}\sum_{i = j}^{m}a_{ij}^2}$. Dear fang-jh, I was wondering if there was a typo in the denominator in this term: $ \sum_{i = 1}^{n}\sum_{i = j}^{m}a_{ij}^2$ .. i guess its a valid sum, but then the (inner) sum should be written from $ j = i$ to $ j = m$, not from $ i = j$ to $ i = m$ ; and the (outer) sum from $ i = 1$ to $ m$, not $ i = 1$ to $ n$ (since if $ n \geq i \geq m$, the second sum is empty), and it is quite a strange sum. Perhaps it may be this: $ \sum_{i = 1}^{n}\sum_{j = 1}^{m}a_{ij}^2$? Dear epitomy01, I check and compare with the Chinese version carefully, find no typo.

Fang-jh wrote: For two given positive integers $ m,n > 1$, let $ a_{ij} (i = 1,2,\cdots,n,j = 1,2,\cdots,m)$ be nonnegative real numbers, not all zero, find the maximum and the minimum values of $ f$, where $ f = \frac {n\sum_{i = 1}^{n}(\sum_{j = 1}^{m}a_{ij})^2 + m\sum_{j = 1}^{m}(\sum_{i = 1}^{n}a_{ij})^2}{(\sum_{i = 1}^{n}\sum_{j = 1}^{m}a_{ij})^2 + mn\sum_{i = 1}^{n}\sum_{i = j}^{m}a_{ij}^2}$. A typo exists. It should be $ f = \frac {n\sum_{i = 1}^{n}(\sum_{j = 1}^{m}a_{ij})^2 + m\sum_{j = 1}^{m}(\sum_{i = 1}^{n}a_{ij})^2}{(\sum_{i = 1}^{n}\sum_{j = 1}^{m}a_{ij})^2 + mn\sum_{i = 1}^{n}\sum_{j = 1}^{m}a_{ij}^2}$ See

Attachments:

shalex is right, please moderator correct it. thank a lot!

The minimum value is $ \frac{m+n}{1+mn}$ while the maximum value is $ 1$. Correct ?

The maximum value is $ 1,$ the minimum value is $ \frac {m+n}{mn+min\{m,n\}}$.

Ahh, I realize it now. In my proof, the equality for the minimum value is $ a_{ij}=0 \forall i \not= j$ and $ a_{ii}=\text{const} \forall i=1, 2, ..., min(m,n)$.

couyld anyone post his proof please

yes! Can you post it?

For sake of convenience, denote $ \sum_{j = 1}^{m} a_{ij} = x_i$ $ \sum_{i = 1}^{n} a_{ij} = y_j$ What we have is $ f = \frac {n \sum ( x_i )^2 + m \sum ( y_i )^2 }{(\sum \sum a_{ij} )^2 + mn \sum \sum a_{ij}^2}$ (1) Maximum of $ f$ We shall prove that the maximum is 1. It suffices to show that $ n \sum (x_i )^2 + m \sum ( y_i )^2 \leq (\sum \sum a_{ij} )^2 + mn \sum \sum a_{ij}^2$ Consider the two following sequence of $ m^2 + n^2$ nonnegative real numbers; $ (x_1 , x_1 , \cdots , x_1 , x_2 , \cdots , x_n , y_1 , y_1 , \cdots , y_1 , y_2 , \cdots , y_m )$ (Where each $ x_i$ appears $ n$ times and each $ y_i$ appears $ m$ times) $ ( \sum \sum a_{ij} , a_{11}, a_{12} , \cdots , a_{mn} , 0 , 0 , \cdots , 0 )$ (the 0's exist to make the number of terms equal) It is obvious that the latter sequence majorizes the former sequence (The largest is the largest, and any other term in the latter sequence is smaller in size to an appropriate term in the former sequence) Since the function $ g(x) = x^2$ is convex in the nonnegative reals, applying Karamata inequality, we arrive at the result. (2) Minimum of $ f$ We wish to show that $ f \geq \frac {m + n}{mn + min(m,n)}$. Assume $ m \geq n$, so that $ min(m,n) = n$ For sake of convenience, denote $ \sum (x_i )^2 = A$ $ \sum (y_i )^2 = B$ $ \sum \sum a_{ij}^2 = C$ $ (\sum \sum a_{ij})^2 = D$ What we wish to prove is that $ \frac {nA + mB}{mnC + D} \geq \frac {m + n}{mn + n}$ .. (*) $ (nA + mB)(mn + n) \geq (mnC + D)(m + n)$ By the Cauchy Inequality, we have $ nA \geq D$, or $ (m + n)nA \geq (m + n)D$ .. (1) Since $ \sum (x_i )^2 = \sum (\sum a_{ij} )^2 \geq \sum \sum a_{ij}^2$, we have $ A \geq C$ likewise, we have $ B \geq C$ Therefore, we have that $ n(mn + n)A \geq n(mn + n)C$ .. (2) $ (m^2 n - n^2 )B \geq (m^2 n - n^2 )C$ .. (3) Adding (1), (2), (3) yields (*), the desired result.

My solution for (1) is incorrect - the upper bound solution is nonsensical. I was drowzy when I wrote it =[ Here's a modified version, please edit the post (I don't know why, but I lost my right to do so): Upper bound - it suffices to show, as mentioned, $ n \sum x_i^2 + m \sum y_i^2 \leq (\sum \sum a_{ij} )^2 + mn \sum \sum a_{ij}^2$ Or, equivalently, $ n \sum x_i^2 - (\sum \sum a_{ij})^2 \leq m \sum ( n \sum a_{ij}^2 - \sum y_i^2 )$ Or, equivalently, $ n \sum x_i^2 - (\sum x_i )^2 \leq m \sum (n \sum a_{ij}^2 - \sum y_i^2 )$ (because $ \sum \sum a_{ij} = \sum x_i$) Now, recall the well-known identity $ n \sum a_i^2 - (\sum a_i )^2 = \sum (a_i - a_j )^2$ The inequality becomes $ \sum (x_i - x_j )^2 \leq m \sum ( \sum (a_{ik} - a_{jk})^2 )$ Applying the cauchy inequality in the form $ (x_i - x_j )^2 = (\sum a_{ik} - \sum a_{jk})^2 \leq m \sum (a_{ik} - a_{jk})^2$ and summing them up yields the result.

WI UM DDDD

\[ f = \frac {n\sum_{i = 1}^{n}(\sum_{j = 1}^{m}a_{ij})^2 + m\sum_{j = 1}^{m}(\sum_{i= 1}^{n}a_{ij})^2}{(\sum_{i = 1}^{n}\sum_{j = 1}^{m}a_{ij})^2 + mn\sum_{i = 1}^{n}\sum_{j=1}^{m}a_{ij}^2}. \]We write these 2 identities: \[n\sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij}^2-\sum_{j=1}^{m}(\sum_{i=1}^{n}a_{ij})^2=\sum_{j=1}^{n}(n\sum_{i=1}^{n}a_{ij}^2-(\sum_{i=1}^{n}a_{ij})^2)=\sum_{j=1}^{m}((n-1)\sum_{i=1}^{n}a_{ij}^2-2*\sum_{1\leq k<l \leq n}(a_{kj}*a_{lj}))= \]\[ = \sum_{j=1}^{m}(\sum_{1\leq k<l \leq n}(a_{kj}^2+a_{lj}^2)-2 * \sum_{1\leq k<l \leq n}(a_{kj}*a_{lj}))= \sum_{j=1}^{m} \sum_{1\leq k<l \leq n}(a_{kj}-a_{lj})^2= \sum_{1 \leq k<l \leq n} \sum_{j=1}^{m}(a_{kj}-a_{lj})^2 \]We get 1 equality: \[n\sum_{i=1}^{n} \sum_{j=1}^{m}a_{ij}^2- \sum_{j=1}^{m}(\sum_{i=1}^{n}a_{ij})^2= \sum_{1\leq k<l \leq n} \sum_{j=1}^{m}(a_{kj}-a_{lj})^2 \]\[n\sum_{i=1}^{n}(\sum_{j=1}^{m}a_{ij})^2-(\sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij})^2=(n-1) \sum_{i=1}^{n}( \sum_{j=1}^{m}a_{ij})^2-2* \sum_{1\leq k<l \leq n}(( \sum_{j=1}^{m}a_{kj})*(\sum_{j=1}^{m}a_{lj}))= \]\[= \sum_{1\leq k <l \leq n}((\sum_{j=1}^{m}a_{kj})^2+(\sum_{j=1}^{m}a_{lj})^2)-2* \sum_{1\leq k<l \leq n}((\sum_{j=1}^{m}a_{kj})*(\sum_{j=1}^{m}a_{lj}))= \sum_{1 \leq k<l \leq n}(\sum_{j=1}^{m}(a_{kj}-a_{lj}))^2 \]We get 2 equality: \[ n\sum_{i=1}^{n}(\sum_{j=1}^{m}a_{ij})^2-(\sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij})^2=\sum_{1 \leq k<l \leq n}(\sum_{j=1}^{m}(a_{kj}-a_{lj}))^2 \]Prove that \[ \frac {n\sum_{i = 1}^{n}(\sum_{j = 1}^{m}a_{ij})^2 + m\sum_{j = 1}^{m}(\sum_{i= 1}^{n}a_{ij})^2}{(\sum_{i = 1}^{n}\sum_{j = 1}^{m}a_{ij})^2 + mn\sum_{i = 1}^{n}\sum_{j=1}^{m}a_{ij}^2} \leq 1 \]\[n\sum_{i=1}^{n}(\sum_{j=1}^{m}a_{ij})^2+m\sum_{j=1}^{m}(\sum_{i=1}^{n}a_{ij})^2\leq(\sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij})^2+mn\sum_{i=1}^{n} \sum_{j=1}^{m}a_{ij}^2 \]\[n\sum_{i=1}^{n}(\sum_{j=1}^{m}a_{ij})^2-(\sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij})^2\leq m(n\sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij}^2-\sum_{j=1}^{m}(\sum_{i=1}^{n}a_{ij})^2) \]We use the two previous proven identities. We have: \[\sum_{1\leq k<l\leq n}(\sum_{j=1}^{m}(a_{kj}-a_{lj}))^2\leq m\sum_{1\leq k<l\leq n}\sum_{j=1}^{m}(a_{kj}-a_{lj})^2 \]The last inequality is true because \[m\sum_{j=1}^{m}(a_{kj}-a_{lj})^2\geq(\sum_{j=1}^{m}(a_{kj}-a_{lj}))^2 \]Equality condition in this case: \[a_{kp}-a_{lp}=a_{kq}-a_{lq} \]When $\begin{cases} 1\leq k<l\leq n \\ 1\leq p< q \leq m \end{cases}$ Prove that \[ \frac {n\sum_{i = 1}^{n}(\sum_{j = 1}^{m}a_{ij})^2 + m\sum_{j = 1}^{m}(\sum_{i= 1}^{n}a_{ij})^2}{(\sum_{i = 1}^{n}\sum_{j = 1}^{m}a_{ij})^2 + mn\sum_{i = 1}^{n}\sum_{j=1}^{m}a_{ij}^2}\geq \frac{m+n}{mn+min\{m;n\}} \]Let be \[n \leq m\]The opposite case is considered similarly. Then \[ \frac {n\sum_{i = 1}^{n}(\sum_{j = 1}^{m}a_{ij})^2 + m\sum_{j = 1}^{m}(\sum_{i= 1}^{n}a_{ij})^2}{(\sum_{i = 1}^{n}\sum_{j = 1}^{m}a_{ij})^2 + mn\sum_{i = 1}^{n}\sum_{j=1}^{m}a_{ij}^2}\geq \frac{m+n}{n(m+1)} \]\[n\sum_{i=1}^{n}(\sum_{j=1}^{m}a_{ij})^2+m\sum_{j=1}^{m}(\sum_{i=1}^{n}a_{ij})^2\geq \frac{m+n}{n(m+1)}*(\sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij})^2+ \frac{m(m+n)}{m+1}*\sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij}^2 \]We prove the last inequality: \[n\sum_{i=1}^{n}(\sum_{j=1}^{m}a_{ij})^2+m\sum_{j=1}^{m}(\sum_{i=1}^{n}a_{ij})^2= \frac{m+n}{m+1}*\sum_{i=1}^{n}(\sum_{j=1}^{m}a_{ij})^2+ \frac{m(n-1)}{m+1}*\sum_{i=1}^{n}(\sum_{j=1}^{m}a_{ij})^2+m\sum_{j=1}^{m}(\sum_{i=1}^{n}a_{ij})^2\geq \]\[\geq \frac{m+n}{n(m+1)}*(\sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij})^2+(\frac{m(n-1)}{m+1}+m)*\sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij}^2=\frac{m+n}{n(m+1)}*(\sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij})^2+\frac{m(m+n)}{m+1}*\sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij}^2 \]Equality condition in this case: $\begin{cases} \sum_{j=1}^{m}a_{kj}=\sum_{j=1}^{m}a_{lj} \quad if \quad n\leq m \\ \sum_{i=1}^{n}a_{ip}=\sum_{i=1}^{n}a_{iq} \quad if \quad m \leq n \\a_{kp}*a_{lp}=0 \\ a_{kp}*a_{kq}=0 \end{cases}$ When $\begin{cases} 1\leq k<l \leq n \\1\leq p<q \leq m \end{cases}$ Answer: \[\frac{m+n}{mn+min\{m;n\}}\leq f \leq 1 \]

Let $c_i=\sum\limits_{j=1}^ma_{ij},1\leq i\leq n,r_j=\sum\limits_{i=1}^na_{ij},1\leq j\leq m$. 1 we find the maximum value of f. For $1\leq i\leq n,1\leq j\leq m$, let $a_{ij}=1$, then $f=1\Rightarrow f_{\max}\geqslant 1$. We now prove $f\leqslant 1$. $f\leqslant 1\Leftrightarrow n\sum\limits_{i=1}^nc_i^2+m\sum\limits_{j=1}^mr_j^2\leqslant \left(\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}a_{ij}\right)^2+mn\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}a_{ij}^2$ $\Leftrightarrow n\sum\limits_{i=1}^n\sum\limits_{1\leqslant j_1<j_2\leqslant m}\left(a_{ij_1}-a_{ij_2}\right)^2\geqslant\sum\limits_{1\leqslant j_1<j_2\leqslant m}\left(r_{j_1}-r_{j_2}\right)^2=\sum\limits_{1\leqslant j_1<j_2\leqslant m}\sum\limits_{i = 1}^{n}\left(a_{ij_1}-a_{ij_2}\right)$ $\Leftrightarrow\sum\limits_{1\leqslant j_1<j_2\leqslant m}\left[\sum\limits_{1\leqslant i_1<i_2\leqslant n}\left(a_{i_1j_1}-a_{i_1j_2}-a_{i_2j_1}+a_{i_2j_2}\right)^2\right]\geqslant 0$ which is obviously true. 2 we find the minimum value of f. Let $m=\min\{m,n\}$. For $1\leq j\leq m$, let $a_{jj}=1$, otherwise $a_{ij}=0$, them $f=\frac{m+n}{m+mn}$. We now prove $f\geqslant\frac{m+n}{m+mn}$. $f\geqslant\frac{m+n}{m+mn}\Leftrightarrow (m+mn)\left(n\sum\limits_{i=1}^nc_i^2+m\sum\limits_{j=1}^mr_j^2\right)\leqslant (m+n)\left[\left(\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}a_{ij}\right)^2+mn\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}a_{ij}^2\right]$ Cauchy-Schwarz inequality $\Rightarrow n\sum\limits_{i=1}^nc_i^2\geqslant\left(\sum\limits_{i=1}^nc_i\right)^2=\left(\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}a_{ij}\right)^2,m\sum\limits_{j=1}^mr_j^2\geqslant\left(\sum\limits_{j=1}^mr_j\right)^2=\left(\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}a_{ij}\right)^2$ Let $A=n\sum\limits_{i=1}^nc_i^2,B=m\sum\limits_{j=1}^mr_j^2,C=\left(\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}a_{ij}\right)^2,D=\sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{m}a_{ij}^2.$ We have $A\geqslant C,B\geqslant C,A\geqslant nD,B\geqslant mD$. $\Leftrightarrow (m+mn)(A+B)\geqslant (m+n)(C+mnD)$ $LHS=(m+mn)A+(m+n)B+(m-1)nB\geqslant (m+mn)nD+(m+n)C+(m-1)nmD=RHS$