What is the minimum value of $f(x)$ with positive real numbers $p,q,r$?

[anemone]

For positive real numbers $p,\,q,\,r$, determine the minimum of the function $f(x)=\sqrt{p^2+x^2}+\sqrt{(q-x)^2+r^2}$.

 

[anemone]

For positive real numbers $p,\,q,\,r$, determine the minimum of the function $f(x)=\sqrt{p^2+x^2}+\sqrt{(q-x)^2+r^2}$.

Hint:

Spoiler

The proposed solution uses the geometry approach that solved it neatly and nicely.:)

Further hint:

Spoiler

Minimum $f(x)$ is $\sqrt{(p+r)^2+q^2}$.

 

[anemone]

Solution of other:

Spoiler

View attachment 4563

Let $AD=p,\,AE=q,\,EC=r$. Let $B$ be a point on $AE$ and let $x=AB$, so that $BE=q-x$. Then $f(x)=DB+BC$. To minimize $DB+BC$, we use the method of reflection. Let $C'$ be the reflection of $C$ in the line $AE$.

Since triangles $BEC$ and $BEC'$ are congruent, $BC=BC'$ and $f(x)=DB+BC'$.

As $x$ varies, $B$ changes its position. But the distance $DB+BC'$ will be a minimum when $B$ lies on the line $DC'$ (as shown in the orange line).

The minimum value of $f(x)$ is then $DB+BC'=DC'$. Let $DD'$ be the perpendicular from $D$ to the line $CC'$. From the right triangle DD'C',

$f(x)_{\text{minimum}}=DC'=\sqrt{DD'^2+D'C'^2}=\sqrt{q^2+(p+r)^2}$