China Mathematical Olympiad

As $$\begin{array}{rl}\min \{f(x_i),f(x_j)\}& =\min \{(x_i+a)(x_i+b),(x_j+a)(x_j+b)\}\\ & \le \sqrt{(x_i+a)(x_i+b)(x_j+a)(x_j+b)}\\ & \le \frac{1}{2}((x_i+a)(x_j+b)+(x_i+b)(x_j+a))\\ & =x_i{x}_j+\frac{1}{2}(x_i+x_j)(a+b)+ab,\end{array}$$ so $$\begin{array}{rl}F& \le \sum _{1\le i<j\le n}{x}_i{x}_j+\frac{a+b}{2}\sum _{1\le i<j\le n}(x_i+x_j)+C_n^2\cdot ab\\ & =\frac{1}{2}\left[{\left(\sum _{i=1}^n{x}_i\right)}^2-\sum _{i=1}^n{x}_i^2\right]+\frac{a+b}{2}(n-1)\sum _{i=1}^n{x}_i+C_n^2\cdot ab\\ & =\frac{1}{2}\left(1-\sum _{i=1}^n{x}_i^2\right)+\frac{n-1}{2}(a+b)+C_n^2\cdot ab\\ & \le \frac{1}{2}\left(1-\frac{1}{n}{\left(\sum _{i=1}^n{x}_i\right)}^2\right)+\frac{n-1}{2}(a+b)+C_n^2\cdot ab\\ & =\frac{1}{2}\left(1-\frac{1}{n}\right)+\frac{n-1}{2}(a+b)+\frac{n(n-1)}{2}ab\\ & =\frac{n-1}{2}\left(\frac{1}{n}+a+b+nab\right).\end{array}$$ The equality holds when $x_1=x_2=\cdots =x_n=\frac{1}{n}$. So the maximum of $F$ is $\frac{n-1}{2}\left(\frac{1}{n}+a+b+nab\right)$.

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Alternative solution.

We show that the maximum value is attained when $x_1=x_2=\cdots =x_n=\frac{1}{n}$, and $F_{\max }=\frac{n-1}{2}\left(\frac{1}{n}+a+b+nab\right)$. We induct on $n$ to show a more general statement: for non-negative real numbers $x_1,x_2,\dots ,x_n$ satisfying $x_1+x_2+\cdots +x_n=s$ (where $s$ is a fixed non-negative real number), the maximum value of $F={\sum }_{1\le i<j\le n}\min \{f(x_i),f(x_j)\}$ is attained when $x_1=x_2=\cdots =x_n=\frac{s}{n}$. Since $F$ is symmetric, we may assume $x_1\le x_2\le \cdots \le x_n$. Note that $f(x)$ is strictly increasing on non-negative real numbers, we have $$F=(n-1)f(x_1)+(n-2)f(x_2)+\cdots +f(x_{n-1}).$$ When $n=2$, $F=f(x_1)\le f(\frac{s}{2})$, equality holds when $x_1=x_2$. Assume that the statement holds for $n$, consider the case of $n+1$. Applying inductive hypothesis on $x_2+x_3+\cdots +x_{n+1}=s-x_1$, we have $$F\le nf(x_1)+\frac{1}{2}n(n-1)f(\frac{s-x_1}{n})=g(x_1),$$ where $g(x)$ is a quadratic function of $x$, the leading coefficient is $1+\frac{n-1}{2n^2}$, and the coefficient of $x$ is $a+b-\frac{n-1}{2n}(a+b+\frac{s}{2n})$, therefore, the axis of symmetry is $$\frac{\frac{n-1}{2n}\left(a+b+\frac{s}{2n}\right)-a-b}{2+\frac{n-1}{n^2}}\le \frac{s}{2(n+1)}.$$ (The above inequality is equivalent to $[(n-1)s-2n(n+1)(a+b)](n+1)\le 2s(2n^2+n-1)$; obviously, left-hand side $<(n^2-1)s<$ right-hand side.) Therefore, $g(\frac{s}{n+1})$ is the maximum of $g(x)$ on $[0,\frac{s}{n+1}]$. Thus, $F$ attains its maximum when $x_2=x_3=\cdots =x_{n+1}=\frac{s-x_1}{n}=\frac{s}{n+1}=x_1$, completing the solution. $\square$