THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

Denoting $y_1=x_1$, $y_{k+1}=x_{k+1}-x_k$, for $k\in \{1,2,\dots ,n-1\}$, we get $x_k=y_1+y_2+\cdots +y_k$, $\text{cu }{y}_1>0,y_k\ge 0,\forall k=\overline{2,n}$. With these notations, the inequality becomes $(a_1+a_2+\cdots +a_n-c(b_1+b_2+\cdots +b_n))y_1+(a_2+a_3+\cdots +a_n-c(b_2+b_3+\cdots +b_n))y_2+\cdots +(a_n-c{b}_n)y_n\ge 0$ ($\ast$).

Plugging $y_1=1,y_2=\cdots =y_n=0$, shows that necessarily $c\le \frac{a_1+a_2+\cdots +a_n}{b_1+b_2+\cdots +b_n}$.

We show that, conversely, for every $c\le \frac{a_1+a_2+\cdots +a_n}{b_1+b_2+\cdots +b_n}$ the inequality under consideration is true, so $\frac{a_1+a_2+\cdots +a_n}{b_1+b_2+\cdots +b_n}$ is the required maximum.

We notice that $c\le \frac{a_1+a_2+\cdots +a_n}{b_1+b_2+\cdots +b_n}\le \frac{a_2+a_3+\cdots +a_n}{b_2+b_3+\cdots +b_n}\le \cdots \le \frac{a_n}{b_n}$. Indeed, the inequality $\frac{a_k+a_{k+1}+\cdots +a_n}{b_k+b_{k+1}+\cdots +b_n}\le \frac{a_{k+1}+\cdots +a_n}{b_{k+1}+\cdots +b_n}$ can be written $a_k(b_{k+1}+\cdots +b_n)\le b_k(a_{k+1}+\cdots +a_n)$ and is justified by the inequalities $\frac{a_k}{b_k}\le \frac{a_{k+1}}{b_{k+1}},\frac{a_k}{b_k}\le \frac{a_{k+2}}{b_{k+2}},\dots ,\frac{a_k}{b_k}\le \frac{a_n}{b_n}$.

This shows that every term of the left member of $(\ast )$ is at least 0, so $(\ast )$ is true.