IMO Team Selection Contest

Applying AM-GM gives $$\begin{array}{rl}\left(\sum _{k=1}^n\frac{x_k}{k}\right)\left(\sum _{k=1}^{n}k{x}_k\right)& =\frac{1}{n}\cdot \left(\sum _{k=1}^n\frac{n{x}_k}{k}\right)\left(\sum _{k=1}^{n}k{x}_k\right)\le \\ & \le \frac{1}{n}\cdot \frac{1}{4}{\left(\sum _{k=1}^n\frac{n{x}_k}{k}+\sum _{k=1}^{n}k{x}_k\right)}^2=\\ & =\frac{1}{4n}{\left(\sum _{k=1}^n{x}_k\left(\frac{n}{k}+k\right)\right)}^2\le \frac{(n+1{)}^2}{4n}{\left(\sum _{k=1}^n{x}_k\right)}^2.\end{array}$$ (The last inequality is proved by $\frac{n}{k}+k\le n+1$, as it is equivalent to $(n-k)(k-1)\ge 0$.) This gives us the necessary upper bound; this bound is achieved for instance if $x_1=x_n=1$ and $x_2=\cdots =x_{n-1}=0$.

For the lower bound, estimate the numerator by Cauchy-Schwarz inequality: $$\left(\sum _{k=1}^n\frac{x_k}{k}\right)\left(\sum _{k=1}^{n}k{x}_k\right)\ge {\left(\sum _{k=1}^n\sqrt{\frac{x_k}{k}}\cdot \sqrt{k{x}_k}\right)}^2={\left(\sum _{k=1}^n{x}_k\right)}^2;$$ the equality holds here if exactly one of $x_i$s is non-zero.