Let $a=x_1\leq x_2\leq ...\leq x_n=b$. Prove the following inequality: $(x_1+x_2+...+x_n )(\frac{1}{x_1} +\frac{1}{x_2} +...+\frac{1}{x_n} )\leq \frac{(a+b)}{4ab} n^2$.