<s>[INST] 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 $$ x=\frac{40}{23}(12-5\sqrt{3}).$$
 [/INST] 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 $b(x)$. $\quad \Box$
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