Minimal value of $\sqrt{(x-a)^2 + b^2} +\sqrt{(x-c)^2 + d^2}$ without derivatives and without distance formula

Question

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.

Answer 1

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. $g(x) = 0$.

Answer 2

Applying Fermat's Principle on reflection:

Assume $a<c$,

\begin{align*} T(x) &= \sqrt{(x-a)^2+b^2}+\sqrt{(x-c)^2+d^2} \\ T'(x) &=\frac{x-a}{\sqrt{(x-a)^2+b^2}}-\frac{c-x}{\sqrt{(x-c)^2+d^2}} \\ 0 &= \sin i-\sin r \\ i &= r \\ \end{align*}

If you've accepted the law of reflection, then

\begin{align*} \frac{x-a}{b} &= \frac{c-x}{d} \\ x &= \frac{ad+bc}{b+d} \end{align*}

where $b,d>0$

See another answer for the case of refraction.