Just make the substitution $y=\frac{1}{x}$ and after expanding the inequality will reduce to showing $(x-1)^2 \ ge 0$ which is trivially true. Equality holds when $x=y=1$.

RockmanEX3 wrote: Prove that all real numbers $x \ne -1$, $y \ne -1$ with $xy = 1$ satisfy the following inequality: $$\left(\frac{2+x}{1+x}\right)^2 + \left(\frac{2+y}{1+y}\right)^2 \ge \frac92$$ (Karl Czakler) We have $$\left(\frac{2+x}{1+x}\right)^2 + \left(\frac{2+y}{1+y}\right)^2 =\left(\frac{2xy+x}{xy+x}\right)^2 + \left(\frac{2+y}{1+y}\right)^2= \left(\frac{2y+1}{1+y}\right)^2 + \left(\frac{2+y}{1+y}\right)^2 =\frac{5y^2+8y+5}{y^2+2y+1}\geq \frac 9 2,$$which transforms as $(y-1)^2\geq 0$.

We can also use AM GM repeatedly till the end result is $\geq \frac{9}{2}$

First, we make the substitution $x=\frac{1}{y}.$ Expanding, we have $$\left(\frac{2+\frac{1}{y}}{1+\frac{1}{y}}\right)^2 + \left(\frac{2+y}{1+y}\right)^2 \ge \frac92$$ $$\frac{4+\frac{4}{y}+\frac{1}{y^2}}{1+\frac{2}{y}+\frac{1}{y^2}}+\frac{4+4y+y^2}{1+2y+y^2} \ge \frac92$$ $$\frac{4y^2+4y+1}{y^2+2y+1}+\frac{y^2+4y+4}{y^2+2y+1} \ge \frac92$$ $$\frac{5y^2+8y+5}{y^2+2y+1} \ge \frac92$$ $$10y^2+16y+10 \ge 9y^2+18y+9$$ $$y^2-2y+1 \ge 0$$ $$(y-1)^2 \ge 0.$$ This is true for all real numbers by the Trivial Inequality, and we are done $\blacksquare$