China Mathematical Competition (Complementary Test)

For any $k=1,2,\dots ,m$, assuming there are $x_k$ students failing to answer the $k$th question correctly, then there are $n-x_k$ ones who answer it correctly and each gets $x_k$ marks from it accordingly. Suppose the sum of the $n$ students' total marks is $S$. Then we have $$\sum _{i=1}^n{p}_i=S=\sum _{k=1}^m{x}_k(n-x_k)=n\sum _{k=1}^m{x}_k-\sum _{k=1}^m{x}_k^2.$$ As each student gets at most $x_k$ marks from the $k$th question, we have $p_1\le {\sum }_{k=1}^m{x}_k.$ Since $p_2\ge \cdots \ge p_n$, then $p_n\le \frac{p_2+p_3+\cdots +p_n}{n-1}=\frac{S-p_1}{n-1}$. Therefore, $$\begin{array}{rl}{p}_1+p_n& \le p_1+\frac{S-p_1}{n-1}=\frac{n-2}{n-1}{p}_1+\frac{S}{n-1}\\ & \le \frac{n-2}{n-1}\cdot \sum _{k=1}^m{x}_k+\frac{1}{n-1}\cdot \left(n\sum _{k=1}^m{x}_k-\sum _{k=1}^m{x}_k^2\right)\\ & =2\sum _{k=1}^m{x}_k-\frac{1}{n-1}\cdot \sum _{k=1}^m{x}_k^2.\end{array}$$ By the Cauchy Inequality, we have $$\sum _{k=1}^m{x}_k^2\ge \frac{1}{m}{\left(\sum _{k=1}^m{x}_k\right)}^2.$$ Then $$\begin{array}{rl}{p}_1+p_n& \le 2\sum _{k=1}^m{x}_k-\frac{1}{m(n-1)}\cdot {\left(\sum _{k=1}^m{x}_k\right)}^2\\ & =-\frac{1}{m(n-1)}\cdot {\left(\sum _{k=1}^m{x}_k-m(n-1)\right)}^2+m(n-1)\\ & \le m(n-1).\end{array}$$ On the other hand, if there is a student who answers all the questions correctly, while the other $n-1$ students fail to answer any questions, then we have $$p_1+p_n=p_1=\sum _{k=1}^m(n-1)=m(n-1).$$ Therefore, the maximum possible value of $p_1+p_n$ is $m(n-1)$. ​