Find the minimum of the function

$$
f(x, y)=\sqrt{(x+1)^{2}+(2 y+1)^{2}}+\sqrt{(2 x+1)^{2}+(3 y+1)^{2}}+\sqrt{(3 x-4)^{2}+(5 y-6)^{2}} \text {, }
$$

defined for all real $x, y>0$.
Note that $\sqrt{(x+1)^{2}+(2 y+1)^{2}}$ is the distance in the coordinate plane from $(0,0)$ to $(x+1,2 y+1) ; \sqrt{(2 x+1)^{2}+(3 y+1)^{2}}$ is the distance from $(x+1,2 y+1)$ to $(3 x+2,5 y+2)$; $\sqrt{(3 x-4)^{2}+(5 y-6)^{2}}$ is the distance from $(3 x+2,5 y+2)$ to $(6,8)$. Thus, $f(x, y)$ is the distance along the polygonal path from $(0,0)$ to $(x+1,2 y+1)$ to $(3 x+2,5 y+2)$ to $(6,8)$, which, by the triangle inequality, cannot be less than the distance from $(0,0)$ to $(6,8)$. Thus $f(x, y) \geq 10$. One then calculates $f(1 / 4,1 / 3)=10$. So 10 is in fact attainable, and it is our minimum.