Find minimal value $ \sqrt {{x}^{2}-5\,x+25}+\sqrt {{x}^{2}-12\,\sqrt {3}x+144}$ without derivatives. Find minimal value  of $ \sqrt {{x}^{2}-5\,x+25}+\sqrt {{x}^{2}-12\,\sqrt {3}x+144}$ without using the derivatives and   without the formula for the distance between  two points.
By using the derivatives I have  found  that the minimal value  is $13$  at $$ x=\frac{40}{23}(12-5\sqrt{3}).$$
 A: If we complete the squares we get $$\sqrt{\left(x-{5\over 2}\right)^2 +{75\over 4}} + \sqrt{ (x-6\sqrt{3})^2 +36}.$$  This is the distance from the point $P(x,0)$ to the point $Q(5/2, -5\sqrt{3}/2)$ plus the distance from $P(x,0)$ to the point $R(6\sqrt{3},6).$  Choosing $x$ amounts to sliding $P$ along the $x$-axis.  The shortest sum of distances will be the straight line segment from $Q$ to $R$.  This length is $$\sqrt{ \left( 6+{5\sqrt{3}\over 2}\right)^2 +(6\sqrt{3}-5/2)^2}=13.$$  
A: Note that the expression in question, $f(x) = \sqrt {{x}^{2}-5\,x+25}+\sqrt {{x}^{2}-12\,\sqrt {3}x+144}$, can be written as follows:
$$
f(x) = \sqrt {(a(x))^2 + (b(x))^2}+\sqrt {(a(x))^2 + (13 - b(x))^2}
$$ 
where 
$$a(x) =\frac{x (- 5\sqrt3 - 12) + 120}{26}  $$
$$b(x) =\frac{x (12\sqrt3 -5) + 50}{26}  $$
Now from $(a(x))^2 \ge 0$ follows
$$
f(x) \ge \sqrt {(b(x))^2}+\sqrt { (13 - b(x))^2} = 13
$$
but also, for $a(x) = 0$ we have $f(x) = 13$. So indeed $13$ is the smallest value that $f(x)$  can attain. Again, this value is  reached for $a(x) = 0$, i.e. $x = \frac{120}{5\sqrt3 + 12} = \frac{40}{23}(12-5\sqrt{3})$ as already given by the OP. This completes the proof. 
I can give a general construction method of how to arrive at the functions $a(x)$ and $b(x)$. $\quad \Box$
