Let $M=\{a_1,a_2,...,a_n\}$ and $f(a_i)=b_i$ for all $1\le i\le n$. Then $\sum_{1\le i<j\le n}a_ia_j\le \sum_{1\le i<j\le n}b_ib_j= \sum_{1\le i<j\le n }a_ia_j$.So $a_ia_j=b_ib_j$, for all $i\ne j$. And it means that $c=\frac{b_i}{a_i}$ for all $i$, where $c$ is constant. So $\sum a_i=\sum b_i=c\cdot \sum a_i$. Since $f$ is not identity, $c\ne 1$. So we get that $\sum a_i=0$ as desired.

oVlad wrote: A finite set $M$ of real numbers has the following properties: $M$ has at least $4$ elements, and there exists a bijective function $f:M\to M$, different from the identity, such that $ab\leq f(a)f(b)$ for all $a\neq b\in M.$ Prove that the sum of the elements of $M$ is $0.$ Let \(a_1,\hdots, a_n\) be the elements of \(M\). Note that \[a_ia_j\leq f(a_i)f(a_j)\]for all \(1\leq i,j\leq n\) and also as \(f\) is a bijection, \[\lvert a_1\cdots a_n\rvert =\lvert f(a_1)\cdots f(a_n)\rvert\]And also note that multiplying all the inequalities above gives us that \[\lvert a_1\cdots a_n\rvert\leq\lvert f(a_1)\cdots f(a_n)\rvert\]so, equality must hold everywhere. Therefore, \(f(a_i)f(a_j)=a_ia_j\) or \(f(a_i)=aa_j\) for some constant \(a\). Therefore, \[\sum_{i=1}^na_i=\sum_{i=1}^nf(a_i)=a\sum_{i=1}^na_i\]and as \(a\neq1\), we see that \(\sum_{i=1}^na_i=0\) as requested.

rama1728 wrote: oVlad wrote: A finite set $M$ of real numbers has the following properties: $M$ has at least $4$ elements, and there exists a bijective function $f:M\to M$, different from the identity, such that $ab\leq f(a)f(b)$ for all $a\neq b\in M.$ Prove that the sum of the elements of $M$ is $0.$ Therefore, \(f(a_i)f(a_j)=a_ia_j\) or \(f(a_i)=aa_j\) for some constant \(a\). How did you know $a$ will be constant? If so then we can take $aa_k=f(a_i)=aa_j$ which means $a=0$?

sanyalarnab wrote: rama1728 wrote: oVlad wrote: A finite set $M$ of real numbers has the following properties: $M$ has at least $4$ elements, and there exists a bijective function $f:M\to M$, different from the identity, such that $ab\leq f(a)f(b)$ for all $a\neq b\in M.$ Prove that the sum of the elements of $M$ is $0.$ Therefore, \(f(a_i)f(a_j)=a_ia_j\) or \(f(a_i)=aa_j\) for some constant \(a\). How did you know $a$ will be constant? If so then we can take $aa_k=f(a_i)=aa_j$ which means $a=0$? Look we have \(f(a_i)f(a_j)=a_ia_j\) for all \(i,j\) so \(a\) is constant???

@rama1728 you made a typo, no need to get upset. What @sanyalarnab is trying to point out is that you said $f(a_i)=aa_j,$ and the indices are messed up. (you used $j$ instead of $i$ in the RHS) rama1728 wrote: Therefore, \(f(a_i)f(a_j)=a_ia_j\) or \(f(a_i)=aa_j\) for some constant \(a\).

oVlad wrote: @rama1728 you made a typo, no need to get upset. What @sanyalarnab is trying to point out is that you said $f(a_i)=aa_j,$ and the indices are messed up. (you used $j$ instead of $i$ in the RHS) rama1728 wrote: Therefore, \(f(a_i)f(a_j)=a_ia_j\) or \(f(a_i)=aa_j\) for some constant \(a\). Ohh ok ok thanks

oVlad wrote: A finite set $M$ of real numbers has the following properties: $M$ has at least $4$ elements, and there exists a bijective function $f:M\to M$, different from the identity, such that $ab\leq f(a)f(b)$ for all $a\neq b\in M.$ Prove that the sum of the elements of $M$ is $0.$ (Not very sure if this is right, but....) Borrowing notation from the above solutions, let $a_i$ be the elements of $M$. Now $\sum_{i=1}^{n} {a_i}^2 = \sum_{i=1}^{n} {b_i}^2$, and $\sum_{i=1}^{n} {a_i} = \sum_{i=1}^{n} {b_i}$. Squaring the second equation and subtracting it from the first, we get that $\sum_ {1 \leq i < j \leq n} a_ia_j= \sum_ {1 \leq i < j \leq n} b_ib_j$. Now it is given that $a_ia_j \leq b_ib_j$, for all $ i \neq j$. Summing up all such inequalities, and using the fact the sum taken twice at a time is equal, equality must hold in all inequalities. So $a_ia_j =b_ib_j$ for all $i \neq j$. Now consider these $3$ equations: \begin{align*} a_ia_j =b_ib_j\\ a_ja_k =b_jb_k\\ a_ia_k =b_ib_k\\ \end{align*}So multiplying first 2 and using third gives that ${a_j}^2 = {b_j}^2$. So $a_j = \pm b_j$ for all $j$.

Claim

Proof

So we get that $a_k= -b_k$, for all $k$. Summing it up, and using that $\sum_{i=1}^{n} {a_i} = \sum_{i=1}^{n} {b_i}$, we get that $\sum_{i=1}^{n} {a_i}=0$, and we're done

Let $M={a_1,a_2,...,a_n}$ and $f(a_i) = x_i$. we have $\sum a_ia_j \le \sum x_ix_j$ but $\sum x_ix_j = \sum a_ia_j$ cause $f$ is bijective so now for $i,j$ we have $a_ia_j = x_ix_j$. Let assume $a_ia_j = x_ix_j$,$a_ja_k = x_jx_k$ and $a_ka_i = x_kx_i$ by multiplying $1,2$ and dividing $3$ we have $a_i = \pm x_i$ but if $a_i = x_i$ then $f$ isn't different from the identity so we have $f(a_i) = -a_i$ and since $f$ is bijective then $\sum a_i = 0$.