calculus
I saw this post but it was too technical for me.
I am looking for proof similar to what is shown in this Quora post by Nicholas Halderman. The part I found confusing/ hard of the proof was where he proves the lemma:'a small displacement in this direction along this curve will result in a point closer to B.', I am looking for alternate geometric proof/ a more elaborate explanation on what he has done in the post.
To answer, I assume $F(x)$ and $G(x)$ are continuous in $I \in \mathbb{R}, I = [a,b]$. For simplicity, let $F(x)=f, G(x) = g, H(x) = h = f-g$. By symmetry I can assume $f \ge g$. Now consider the function $h(x)$:
Suppose $a(s),b(t)$ are two curves in $\mathbb R^2$ with parameter interval $(0,1).$ Assume $a'(s),b'(t)$ never vanish, otherwise normal vectors make no sense. Define
$$f(s,t)= |a(s)-b(t)|^2.$$
Suppose $f(s_0,t_0)>0$ is the minimum value of $f$ (hence $\sqrt {f(s_0,t_0)}$ is the distance between the two curves). Then $(s_0,t_0)$ is a critical point of $f.$ Thus
$$\frac{\partial f}{\partial s}(s_0,t_0) = 2a'(s_0)\cdot(a(s_0)-b(t_0)) = 0$$
and
$$\frac{\partial f}{\partial t}(s_0,t_0) = 2b'(t_0)\cdot(b(t_0)-a(s_0)) = 0.$$
(Here $\cdot$ denotes the dot product.) Thus both $a'(s_0),b'(t_0)$ are perpendicular to the vector $b(t_0)-a(s_0).$ Thus the vector $b(t_0)-a(s_0)$ is perpendicular to $a$ at $a(s_0),$ and is perpendicular to $b$ at $b(t_0).$ This is the desired conclusion.
Best Answer
A geometric argument, assuming that a shortest segment exists:
Take any point $A$ on the first curve $\gamma_1$. In order to find the point nearest to $A$ on the second curve $\gamma_2$ draw a small circle centered at $A$ and blow it up until it meets, in fact touches, the second curve $\gamma_2$ at some point $A'$. From all points on $\gamma_2$ this point $A'$ is nearest to $A$. The segment $A A'$ is a radius of the circle, hence orthogonal to the common tangent of the circle and $\gamma_2$ at $A'$.
Of course this works also in the other direction. It follows that for any segment from $\gamma_1$ to $\gamma_2$ that is not orthogonal on both curves the above circle construction finds a shorter one.