IMC 2000 · Day 1 · P4

a) $$\left( \sum_{i=1}^{n} \frac{x_i}{\sqrt{i}} \right)^2 = \sum_{i,j} \frac{x_i x_j}{\sqrt{i}\sqrt{j}} \ge \sum_{i=1}^{n} \frac{x_i}{\sqrt{i}} \sum_{j=1}^{i} \frac{x_j}{\sqrt{j}} \ge \sum_{i=1}^{n} \frac{x_i}{\sqrt{i}}\, i\, \frac{x_i}{\sqrt{i}} = \sum_{i=1}^{n} x_i^2$$

b) $$\sum_{m=1}^{\infty} \frac{1}{\sqrt{m}} \left( \sum_{i=m}^{\infty} x_i^2 \right)^{1/2} \le \sum_{m=1}^{\infty} \frac{1}{\sqrt{m}} \sum_{i=m}^{\infty} \frac{x_i}{\sqrt{i - m + 1}}$$ by a) $$= \sum_{i=1}^{\infty} x_i \sum_{m=1}^{i} \frac{1}{\sqrt{m}\sqrt{i - m + 1}}$$ You can get a sharp bound on $$\sup_i \sum_{m=1}^{i} \frac{1}{\sqrt{m}\sqrt{i - m + 1}}$$ by checking that it is at most $$\int_0^{i+1} \frac{1}{\sqrt{x}\sqrt{i + 1 - x}}\,dx = \pi$$ Alternatively you can observe that $$\begin{align*} \sum_{m=1}^{i} \frac{1}{\sqrt{m}\sqrt{i + 1 - m}} &= 2 \sum_{m=1}^{i/2} \frac{1}{\sqrt{m}\sqrt{i + 1 - m}} \le \\ &\le 2 \frac{1}{\sqrt{i/2}} \sum_{m=1}^{i/2} \frac{1}{\sqrt{m}} \le 2 \frac{1}{\sqrt{i/2}} \cdot 2 \sqrt{i/2} = 4 \end{align*}$$