if $n$ be give postive integers,and let $a=(a_{1},a_{2},\cdots,a_{n})$,and define $$S(a)=\sum_{i=1}^{n}3^{i-1}a_{i},~~~T(a)=\sum_{i=1}^{n}\dfrac{a_{i}}{3^{i-1}}$$ Assmue $m,k$ be postive integer such $m\ge 2k$,and define $$A=\left\{a=(a_{1},a_{2},\cdots,a_{n})|k=S(a),a_{i}\in Z,|a_{i}|\le m,i=1,2,\cdots,n\right\}$$ show that $$\dfrac{\sum_{a\in A}T(a)}{|A|}\le k$$
This question looks very interesting, he cleverly combined with combination of an inequality problem, this is the last question we test today, finally thinking for half an hour, has not found a breakthrough, I guess this question is the result of an article?If you've seen it before, or if you can find it, I will thank you very much
One can show that your inequality holds true whenever $2k\geqslant 3m$. The given inequality is equivalent to $$\frac{\displaystyle\sum_{a\in A}T(a)}{|A|}\leqslant k\Leftrightarrow \sum_{a\in A}T(a)\leqslant k|A|=\sum_{a\in A}S(a)$$ by definition of the set $A$. Therefore we need to show that $$0\leqslant \sum_{a\in A}S(a)-T(a)$$ Suppose it isn't true. Then there exists some $a^*\in A$ such that $S(a^*)-T(a^*)<0$ or equivalently $$k<T(a^*)=a^*_1+\frac{a^*_2}{3}+...+\frac{a^*_n}{3^{n-1}}\leqslant |a_1|+\frac{|a^*_2|}{3}+...+\frac{|a^*_n|}{3^{n-1}}\leqslant m\Big(1+\frac{1}{3}+...+\frac{1}{3^{n-1}}\Big)$$ The last sum one can bound above by $$1+\frac{1}{3}+...+\frac{1}{3^{n-1}}<\sum_{k=0}^{\infty}\frac{1}{3^k}=\frac{3}{2}$$ hence we get $$k<\frac{3m}{2}\Rightarrow 2k<3m$$ which would raise a contradiction.