This problem is wrong.

HuangZhen wrote: Suppose $S=\{1,2,3,...,2017\}$,for every subset $A$ of $S$,define a real number $f(A)\geq 0$ such that: $(1)$ For any $A,B\subset S$,$f(A\bigcup B)+f(A\bigcap B)\leq f(A)+f(B);$ $(2)$ For any $A\subset B\subset S$,$f(A)\leq f(B);$ $(3)$ For any $k,j\in S$,$$f(\{1,2,...,k+1\})\geq f(\{1,2,...,k\}\bigcup \{j\});$$$(4)$ For empty set $\varnothing$,$f(\varnothing)=0;$ Confirm that for any three-element subset $T$ of $S$,the inequality below holds.$$f(T)\leq \frac{27}{19}f({1,2,3})$$ $$f(T)\leq \frac{27}{19}f(\{1,2,3\})$$

It is said that this problem is a wrong one with no solutions.And many thanks to sqing for pointing out my typing mistake.

HuangZhen wrote: It is said that this problem is a wrong one with no solutions.And many thanks to sqing for pointing out my typing mistake. you are welcome .

HuangZhen wrote: Suppose $S=\{1,2,3,...,2017\}$,for every subset $A$ of $S$,define a real number $f(A)\geq 0$ such that: $(1)$ For any $A,B\subset S$,$f(A\bigcup B)+f(A\bigcap B)\leq f(A)+f(B);$ $(2)$ For any $A\subset B\subset S$,$f(A)\leq f(B);$ $(3)$ For any $k,j\in S$,$$f(\{1,2,...,k+1\})\geq f(\{1,2,...,k\}\bigcup \{j\});$$$(4)$ For empty set $\varnothing$,$f(\varnothing)=0;$ Confirm that for any three-element subset $T$ of $S$,the inequality below holds.$$f(T)\leq \frac{27}{19}f(\{1,2,3\})$$ Is the number $ \frac{27}{19} $ wrong?

Nah, the corrected version is HuangZhen wrote: Suppose $S=\{1,2,3,...,2017\}$,for every subset $A$ of $S$,define a real number $f(A)\geq 0$ such that: $(1)$ For any $A,B\subset S$,$f(A\bigcup B)+f(A\bigcap B)\leq f(A)+f(B);$ $(2)$ For any $A\subset B\subset S$,$f(A)\leq f(B);$ $(3)$ For any $k,j\in S$,$$f(\{1,2,...,k\})\geq f((\{1,2,...,k\}- \{k\})\bigcup \{j\});$$$(4)$ For empty set $\varnothing$,$f(\varnothing)=0;$ Confirm that for any three-element subset $T$ of $S$,the inequality below holds.$$f(T)\leq \frac{27}{19}f(\{1,2,3\})$$ And for general $|T|=m$, $f(T)\le \frac{m^m}{m^m-(m-1)^{m-1}} f(\{1,2,...,m\})$.

smy2012 wrote: And for general $|T|=m$, $f(T)\le \frac{m^m}{m^m-(m-1)^{m}} f(\{1,2,...,m\})$. Isn't this right?

smy2012 wrote: Nah, the corrected version is HuangZhen wrote: Suppose $S=\{1,2,3,...,2017\}$,for every subset $A$ of $S$,define a real number $f(A)\geq 0$ such that: $(1)$ For any $A,B\subset S$,$f(A\bigcup B)+f(A\bigcap B)\leq f(A)+f(B);$ $(2)$ For any $A\subset B\subset S$,$f(A)\leq f(B);$ $(3)$ For any $k,j\in S$,$$f(\{1,2,...,k\})\geq f((\{1,2,...,k\}- \{k\})\bigcup \{j\});$$$(4)$ For empty set $\varnothing$,$f(\varnothing)=0;$ Confirm that for any three-element subset $T$ of $S$,the inequality below holds.$$f(T)\leq \frac{27}{19}f(\{1,2,3\})$$ And for general $|T|=m$, $f(T)\le \frac{m^m}{m^m-(m-1)^{m-1}} f(\{1,2,...,m\})$. How to prove it?

We define $f(x_0;x_0;...;x_0;x_1;x_2;...;x_n)=f(x_0;x_1;...;x_n)$ with $x_0;x_1;...;x_n$ are the elements of $S$ $\forall a,b,c\in S$, we have : $\frac{5}{3}f(a;b;c)$$\leq$$f(a;b;c)+\frac{2}{3}(f(a)+f(b)+f(c))$ $\leq f(a;b;c)+2f(1)$$\leq f(1;a;b;c)+2f(1)$$\leq f(1;a)+f(1;b;c)+f(1)$ $\leq f(1;a)+f(1;b)+f(1;c)\leq 3f(1;2)$ Then, $\frac{19}{9}f(a;b;c)$$\leq f(a;b;c)+2f(1;2)$$\leq f(1;2;a;b;c)+2f(1;2)$ $\leq f(1;2;a)+f(1;2;b;c)+f(1;2)$$\leq f(1;2;a)+f(1;2;b)+f(1;2;c)$ $\leq 3f(1;2;3)$$\Rightarrow f(a;b;c)\leq \frac{27}{19}f(1;2;3)$

f(a)<=f(1) why?

So why RiJongYong wrote: f(a)<=f(1) why?

It was said that during the test, the condition$(4) f(\varnothing)=0$ was omitted due to some printing problem, and the committee failed to notify the contestants in time. Consequently, anyone who successfully constructed a counterexample without $(4)$ was given full points, and those who added $(4)$ and solved the modified problem also got full mark. However, many failed to find that the problem was wrong and got nothing right after wasting several hours on it. The accident might have affected the national team's namelist, some people would say. Then, this version on AoPS has been corrected.

The following sequence of inequalities proves the statement for an arbitrary triple $(abc)$. $$3f(123)\ge f(12a)+f(12b)+f(12c)\ge 2f(12)+ f(12abc)\ge 2f(12)+f(abc)$$$$3f(12)\ge f(1a)+f(1b)+f(1c)\ge 2f(1)+f(abc)$$$$3f(1)\ge f(a)+f(b)+f(c)\ge f(abc)$$Combine the above and get that: $$f(123)\ge \frac{2}{3}f(12)+\frac{1}{3}f(abc)\ge \frac{4}{9}f(1)+\frac{2}{9}f(abc)+\frac{1}{3}f(abc) \ge \frac{4}{27}f(abc)+ \frac{2}{9}f(abc)+ \frac{1}{3}f(abc)=\frac{19}{27}f(abc)$$ We can also get a more general result ,consider an arbitrary $m-$tuple $(a_1a_2...a_m)$ then for any $k>0$ $$f(123...k)\ge \frac{m^k-(m-1)^k}{m^k}f(a_1a_2...a_m)$$ We prove this inductively using this inequality: $$mf(123...k)\ge \sum_{i=1}^m f(123...(k-1)a_i)\ge (m-1)f(123...(k-1))+f(a_1a_2...a_m)$$