China Mathematical Olympiad

The maximum is $n-1$. By homogeneity, we can assume without loss of generality that ${\sum }_{i=1}^n{a}_i={\sum }_{i=1}^n{b}_i=1$.

First, it is clear that if $a_1=1$, $a_2=a_3=\cdots =a_n=0$ and $b_1=0$, $b_2=b_3=\cdots =b_n=\frac{1}{n-1}$, then ${\sum }_{i=1}^n{a}_i(a_i+b_i)=1$, ${\sum }_{i=1}^n{b}_i(a_i+b_i)=\frac{1}{n-1}$, hence $$\frac{{\sum }_{i=1}^n{a}_i(a_i+b_i)}{{\sum }_{i=1}^n{b}_i(a_i+b_i)}=n-1.$$ Now we prove that for any real numbers $a_1,a_2,\dots ,a_n,b_1,b_2,\dots ,b_n$ satisfying ${\sum }_{i=1}^n{a}_i={\sum }_{i=1}^n{b}_i=1$, we have $$\frac{{\sum }_{i=1}^n{a}_i(a_i+b_i)}{{\sum }_{i=1}^n{b}_i(a_i+b_i)}⩽n-1.$$ Note that the denominator is positive, it is equivalent to show that $$\sum _{i=1}^n{a}_i(a_i+b_i)\le (n-1)\sum _{i=1}^n{b}_i(a_i+b_i),$$ i.e., $$(n-1)\sum _{i=1}^n{b}_i^2+(n-2)\sum _{i=1}^n{a}_i{b}_i\ge \sum _{i=1}^n{a}_i^2.$$ By symmetry, we can assume that $b_1$ is the smallest one among $b_1,b_2,\dots ,b_n$. Then $$\begin{array}{rl}& (n-1)\sum _{i=1}^n{b}_i^2+(n-2)\sum _{i=1}^n{a}_i{b}_i\\ & \ge (n-1)b_1^2+(n-1)\sum _{i=2}^n{b}_i^2+(n-2)\sum _{i=1}^n{a}_i{b}_1\\ & \ge (n-1)b_1^2+{\left(\sum _{i=2}^n{b}_i\right)}^2+(n-2)b_1\\ & =(n-1)b_1^2+(1-b_1{)}^2+(n-2)b_1\\ & =n{b}_1^2+(n-4)b_1+1\\ & \ge 1=\sum _{i=1}^n{a}_i\ge \sum _{i=1}^n{a}_i^2.\end{array}$$ $\square$