Find minimal value $ \sqrt {{x}^{2}-5\,x+25}+\sqrt {{x}^{2}-12\,\sqrt {3}x+144}$ without derivatives.

Question

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

Answer 1

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 $b(x)$. $\quad \Box$

Answer 2

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