1. Let $x, y$ be positive real numbers. Find the minimum value of
$$
x+y+\frac{|x-1|}{y}+\frac{|y-1|}{x}
$$

(Gu Bin, Jin Aiguo)
1. Let $f(x, y)=x+y+\frac{|x-1|}{y}+\frac{|y-1|}{x}$.

If $x \geqslant 1, y \geqslant 1$, then
$$
f(x, y) \geqslant x+y \geqslant 2 \text {; }
$$

If $0<x<1, 0<y<1$, then
$$
f(x, y) = x + y + \frac{1-x}{y} + \frac{1-y}{x} = \left(x + \frac{1-x}{y}\right) + \left(y + \frac{1-y}{x}\right) \geqslant 2.
$$

For $0<x<1, y \geqslant 1$, we have
$$
f(x, y) = x + y + \frac{1-x}{y} + \frac{y-1}{x} \geqslant x + y + \frac{1-x}{y} \geqslant 2.
$$

For $x \geqslant 1, 0<y<1$, we have
$$
f(x, y) = x + y + \frac{x-1}{y} + \frac{1-y}{x} \geqslant x + y + \frac{1-y}{x} \geqslant 2.
$$

For $x>0, y>0$, we have $f(x, y) \geqslant 2$. Also, $f(1,1)=2$, hence the minimum value is 2.