Minimal value of $\sqrt{(x-a)^2 + b^2} +\sqrt{(x-c)^2 + d^2}$ without derivatives and without distance formula Let  $f(x) = \sqrt{(x-a)^2 + b^2} +\sqrt{(x-c)^2 + d^2}$, where all coefficients are real. 
It can be shown using a distance formula that the minimal value of $f(x)$ is 
$D = \sqrt{(a-c)^2+(|b|+|d|)^2}$.
Show that result without derivatives and without a distance formula. At what value of $x$ does the minimum of $f(x)$ occur?
Hint: this is a generalization of this question.

Let, w.l.o.g., $b > 0$ and $d > 0$. 
Define 
$$
x_0 = \frac{ad+bc}{b+d}
$$
Further, as already stated in the question, the minimum value of $f$ is
$D = \sqrt{(a-c)^2+(b+d)^2}$.
The following equality holds
$$
f(x) = \sqrt {(g(x))^2 + (h(x))^2}+\sqrt {(g(x))^2 + (D - h(x))^2}
$$ 
where 
$$g(x) = \frac{b+d}{D} \; (x-x_0) $$
and
$$h(x) =\frac{1}{D} \; \Big[(c-a)\, x + a^2 - c a + b^2 + d b\Big] $$
Now from $(g(x))^2 \ge 0$ follows
$$
f(x) \ge \sqrt {(h(x))^2}+\sqrt { (D - h(x))^2} = D
$$
with equality for $g(x) =0$ which corresponds to $x=x_0$. So indeed $D$ is the smallest value that $f(x)$  can attain, and this happens at $x=x_0$. This completes the proof without distance measures and without derivatives.  $\quad \Box$
If required, the formulae can be re-interpreted geometrically. Note that $f(x)$ is the sum of the distances from point $N = (x,0)$ to the points $P_1 = (a,b)$ and $ P_2 = (c,-d)$. Consider the line $P_1 P_2$. From a distance argument, the minimum of $f(x)$ is  the length of $P_1 P_2$ which is $D$.
Define a point $R$ on that line $P_1 P_2$  such that $NR$ is perpendicular to  $P_1 P_2$. Then $g(x)$ is the length of  $NR$,  $h(x)$ is the length of  $P_1R$, and $D-h(x)$ is the length of  $P_2R$. Since $NR$ is perpendicular to  $P_1 P_2$, the two square roots  then calculate the distances $NP_1$ and $NP_2$ in a Pythagorean way. The minimum of $f(x)$ is obtained for $N = R$, i.e. $g(x) = 0$. 
