It Kantorovich inequality,which is Let $t_i>0$ and $\sum_{i=1}^{n}{t_i}=1$ and $a_1\le a_2\le \cdot\cdot\cdot \le a_n$, then \[\sum_{i=1}^{n}{t_ia_i}\sum_{i=1}^{n}{\frac{t_i}{a_i}}\le \frac{(a_1+a_n)^2}{4a_1a_n}\]

I agree that inequality is nice. Here is my proof. Using Am-Gm inequality, we have: \[ \left( \sum_{k=1}^{n} kx_k \right) \left(\sum_{k=1}^{n} \frac{x_k}{k} \right) \leq \frac{1}{4} \left( \frac{1}{\sqrt{n}}\sum_{k=1}^{n} kx_{k}+\sqrt{n}\sum_{k=1}^{n} \frac{x_k}{k} \right)^2 = \frac{1}{4} \left(\sum_{k=1}^{n} \frac{n+k^2}{k\sqrt{n}} x_k \right)^2 \leq \frac{(n+1)^2}{4n} \left(\sum_{k=1}^{n} x_k\right)^2 \] where last inequality holds because $\frac{n+k^2}{k\sqrt{n}}\le \frac{n+1}{\sqrt{n}} \Leftrightarrow (k-1)(k-n)\leq 0$. $\blacksquare$

danepale wrote: Let $n \ge 1$ and $x_1, \ldots, x_n \ge 0$. Prove that $$ (x_1 + \frac{x_2}{2} + \ldots + \frac{x_n}{n}) (x_1 + 2x_2 + \ldots + nx_n) \le \frac{(n+1)^2}{4n} (x_1 + x_2 + \ldots + x_n)^2 .$$ Let us rewrite the inequality as: $\frac{n+1}{2\sqrt{n}}\sum_{i=1}^nx_i\ge2\sqrt{(\sum_{i=1}^nix_i)(\sum_{i=1}^n\frac{x_i}{i}})$ $\Longrightarrow 2\sqrt{(\sum_{i=1}^nix_i)(\sum_{i=1}^n\frac{x_i}{i}})\le\sum_{i=1}^nix_i+\sum_{i=1}^n\frac{x_i}{i}$ LHS$\le\frac{1}{4}(\frac{1}{\sqrt{n}}\sum_{i=1}^nix_i+\sqrt{n}\sum_{i=1}^n\frac{x_i}{i})^2\le\frac{(n+1)^2}{4n}(\sum_{i=1}^nx_i)^2$ Which is true! And we are done!