SAUDI ARABIAN IMO Booklet 2023

Note that for each $k=1,2,\dots ,n$, we have $$\frac{b_1}{a_1}+\frac{b_2}{a_2}+\cdots +\frac{b_k}{a_k}\ge k\sqrt[k]{\frac{b_1}{a_1}\frac{b_2}{a_2}\dots \frac{b_k}{a_k}}\ge k,$$ which means that $$s_k=\frac{b_1-a_1}{a_1}+\frac{b_2-a_2}{a_2}+\cdots +\frac{b_k-a_k}{a_k}=\frac{b_1}{a_1}+\frac{b_2}{a_2}+\cdots +\frac{b_k}{a_k}-k\ge 0.$$ For each $k=1,2,\dots ,n$, we have $$s_k-s_{k-1}=\frac{b_k-a_k}{a_k}⟹b_k-a_k=a_k{s}_k-a_k{s}_{k-1}$$ where $s_0=0$. Hence, $$\begin{array}{rl}& (b_1-a_1)+(b_2-a_2)+\cdots +(b_n-a_n)\\ & =a_1{s}_1-a_2{s}_1+a_2{s}_2-a_3{s}_2+\cdots +a_n{s}_n-a_n{s}_{n-1}\\ & =s_1(a_1-a_2)+s_2(a_2-a_3)+\cdots +s_{n-1}(a_{n-1}-a_n)+s_n{a}_n\ge 0,\end{array}$$ as desired.