14th Turkish Mathematical Olympiad

By the Cauchy-Schwarz inequality, $$\sum _{i\ne j}\frac{a_i}{a_j}\cdot \sum _{i\ne j}{a}_i{a}_j\ge {\left(\sum _{i\ne j}{a}_i\right)}^2={\left((n-1)\sum _{i=1}^n{a}_i\right)}^2=(n-1{)}^2{t}^2.$$ On the other hand, $$\sum _{i\ne j}{a}_i{a}_j={\left(\sum _{i=1}^n{a}_i\right)}^2-\left(\sum _{i=1}^n{a}_i^2\right)=t^2-t.$$ The required inequality follows.