This solution works.

Attachments:
bw1992.pdf (35kb)

Use the Adreescu inequality, and then the basic$ (x+y)^2\geq4xy$

gwen01 wrote: Prove that for any positive $ x_1,x_2,\ldots,x_n,y_1,y_2,\ldots,y_n$ the inequality \[ \sum_{i = 1}^n\frac1{x_iy_i}\ge\frac {4n^2}{\sum_{i = 1}^n(x_i + y_i)^2} \] holds. $ \sum_{i = 1}^n\frac1{x_iy_i}\ge\frac {4n^2}{\sum_{i = 1}^n4x_iy_i} \ge \frac {4n^2}{\sum_{i = 1}^n(x_i + y_i)^2}$ The first by AM-HM, and the second by $ (x_i + y_i)^2 \ge 4x_iy_i$ and using that $ x \to \frac{1}{x}$ is decreasing.

A bit of a revive, but here's a solution using Cauchy-Schwarz. We'll rewrite the given inequality as \[ \sum_{i=1}^n \frac{1}{x_i y_i} \cdot \sum_{i=1}^n (x_i+y_i)^2 \geq 4n^2. \] Now we'll use Cauchy in the first and AM-GM in the second step and get \begin{align*} \sum_{i=1}^n \frac{1}{x_i y_i} \cdot \sum_{i=1}^n (x_i+y_i)^2 &\stackrel{\text{CS}}{\geq} \left(\sum_{i=1}^n \frac{x_i+y_i}{\sqrt{x_iy_i}} \right)^2 \\ &\geq \left(\sum_{i=1}^n \frac{2\sqrt{x_iy_i}}{\sqrt{x_iy_i}} \right)^2 \\ &= (2n)^2 \\ &= 4n^2. \end{align*} Hence, we're done. $\hfill\blacksquare$