Beginners' Competition

Since $xy=1$, we may assume that $x\ne 0$ and $y\ne 0$. By substituting $y=\frac{1}{x}$ we achieve $${\left(\frac{2+x}{1+x}\right)}^2+{\left(\frac{2+y}{1+y}\right)}^2={\left(\frac{2+x}{1+x}\right)}^2+{\left(\frac{2x+1}{x+1}\right)}^2=\frac{5x^2+8x+5}{x^2+2x+1}$$ and it remains to show that $$\frac{5x^2+8x+5}{x^2+2x+1}\ge \frac{9}{2}$$ This inequality is equivalent to the inequality $$(x-1{)}^2\ge 0$$ and hence everything is proved.