Solution: $\sum \frac{x_i}{y_i} = \sum \frac{{x_i}^2}{x_iy_i} \geq \frac{\left ( \sum x_i \right)^2}{\sum x_iy_i}(*)$ and using ipoteses that $\sum x_iy_i \leq \sum x_i$ we get $(*) \geq \frac {\left ( \sum x_i \right)^2}{\sum x_i} = \sum x_i$ . q.e.d. i used this inequality: If $a_i,b_i>0$ then we have: $\sum \frac{{a_i}^2}{b_i} \geq \frac { \left ( \sum a_i \right )^2}{\sum b_i}$ which can be proved easily by using Cauchy-Buniakowski-Schwarz.

Another solution is to assume by contradiction that the inequality you have to show is false.Well,by adding this with the one from the hyphotessis and using the well known inequality $a+\frac{1}{a}\geq2$ we get very easily a contradiction.