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 $$ x=\frac{40}{23}(12-5\sqrt{3}).$$

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 $$\sqrt{ \left( 6+{5\sqrt{3}\over 2}\right)^2 +(6\sqrt{3}-5/2)^2}=13.$$  
