Optimising the given function I am required to find minimum value of the following function without calculus.
$$f(a,b)=\sqrt{a^2+b^2}+2\sqrt{a^2+b^2-2a+1}+\sqrt{a^2+b^2-6a-8b+25}$$
My attempt:
I realised that I can write the function as $$f(a,b)=\sqrt{(a-0)^2+(b-0)^2}+2\sqrt{(a-1)^2+(b-0)^2}+\sqrt{(a-3)^2+(b-4)^2}$$ which when $a$ and $b$ replaced with coordinates in Cartesian plane represents distances from points $(1,0),(0,0)$ and $(3,4).$ But I'm not sure how I can minimise this.
 A: 
Let me give an graphical representation... I mean this is more like an physics-based informal proof. It shows that if you pull the rope out the ring is gonna go down to the lowest point and the rope will align with the dotted lines.
Also you can think of it geometrically: as the former answer said, the length of any side of a triangle is always less than the sum of the other two, so there we goes.
A: This is an instance of the Weber problem: Where to locate a facility such that the total weighted transportation cost to the clients is minimal? In this case, the facility will be located in $(a,b)$ and the locations are in $(0,0)$, $(1,0)$, and $(3,4)$, with weights 1, 2, and 1.
"A historical perspective on location problems" discusses the history of this problem, and mentions iterative algorithms, but also analog approaches. The analog approach is explained in this paper for example:

A map of the area in question is placed on a board, and holes are drilled in the points where the demand locations are. Strings are passed through the holes, and weights proportional to the economic “weights” are hung on them. The other edges of the strings are tied together. It is quite obvious that the stationary situation reached after possibly a few oscillations is the equilibrium point, namely, the solution of the minimization problem. Obviously, the accuracy of the method is limited by the friction of the strings in the holes, and, in fact, it looks quite primitive when the alternative of an efficient numerical procedure is available.

A: Let $X(a,b),O(0,0),A(1,0),B(3,4)$.
Then, as you pointed out, we want to minimize $XO+XA+XA+XB$.

*

*For any $X$, it holds that $XO+XA\geqslant OA$ whose equality is attained when $X$ is on the line segment $OA$.


*For any $X$, it holds that $XA+XB\geqslant AB$ whose equality is attained when $X$ is on the line segment $AB$.
Therefore, for any $X$, it holds that
$$XO+XA+XA+XB\geqslant OA+AB=1+2\sqrt 5$$ whose equality is attained when $X=A$.
A: The function to be minimized is composed of three addends, one of which has a double coefficient. If we eliminate it, imposing that
$\sqrt{a^{2}+b^{2}-2a+1}=0$
give zero contribution, we get:
$a=1$ and $b=0$.
Therefore the minimum is:
$f(1,0)=\sqrt{1^2+0^2}+2\sqrt{1^2+0^2-2+1}+\sqrt{1^2+0^2-6-0+25}=1+\sqrt{20}=1+2\sqrt{5}$.
