Finding the minimum of $f(x,y)$ by simple inspection Let $$f(x,y) = \sqrt{x^2 + y^2}+ \sqrt{x^2 + y^2-2x+1}+\sqrt{x^2 + y^2-2y+1}+\sqrt{x^2 + y^2-6x-8y +25}$$
Find minimum value of $f$.
Now, We see that we can complete the squares, but how do we proceed after that? Is there any general technique for such problems?
 A: Let $d:\mathbb{R}^2 \rightarrow \mathbb{R}$, $ ((a,b),(c,d))\longmapsto\sqrt{(a-c)^2 + (b-d)^2} $ then  $d((a,b),(c,d)) $ means the distance between $(a,b),(c,d)$. So, we can change $f(x,y)$ to the following expression.
$$f(x,y) = d((x,y),(0,0))+ d((x,y),(1,0))+ d((x,y),(0,1))+d((x,y),(3,4))$$

The answer is: $5+\sqrt2$. In case of proof, we can plot it on the two-dimensional plane to use the inequality of triangle.
For any point $\color{purple}{P(x,y)}$ other than $(\frac{3}{7},\frac{4}{7})$,$$f(x,y)=\overline{\color{purple}{P}O}+\overline{\color{purple}{P}A}+\overline{\color{purple}{P}B}+\overline{\color{purple}{P}C}\\=\color{blue}{\underbrace{(\overline{\color{purple}{P}O}+\overline{\color{purple}{P}C})}_{two\,  side\, of\,\triangle O\color{purple}{P}C}}+\color{red}{\underbrace{(\overline{\color{purple}{P}A}+\overline{\color{purple}{P}B})}_{two\,  side\, of\,\triangle A\color{purple}{P}B}}\\>\overline{CO}+\overline{AB}\\=\overline{DO}+\overline{DC}+\overline{AD}+\overline{DB}\\=f(\frac{3}{7},\frac{4}{7})\\=5+\sqrt2$$
A: That would be the intersection of the diagonals of $ABCD$ where $A(0,0),B(0,1),C(3,4),D(1,0)$ so it's point $(3/7,4/7)$ and $f(3/7,4/7)=\sqrt 2+5$

A: Once you complete the squares and find that the four components are centered in
$$
\left( {0,0} \right),\left( {1,0} \right),\left( {0,1} \right),\left( {0,1} \right),\left( {3,4} \right)
$$
the next step is to realize that
$$
\sqrt {x^{\,2}  + y^{\,2} }  = r
$$
that is that the value of each component corresponds to 
the distance from the center point (they are vertical cones).
So the sum will be equal to the sum of the distances from the four points,
which is minimum at ...
