I am reading Differential Geometry: A First Course in Curves and Surfaces and in Example 1a it asked me to prove that if $\alpha(t)$ is a function that goes through two points, then prove the line segment gives the smallest length. To prove this, I started with this: $$s=\int_a^b\sqrt{1+\alpha'(t)^2}dt$$To minimize $s$, we should minimize the integrand. So: $$\frac{d}{dt}\sqrt{1+\alpha'(t)^2}=\frac{\alpha'(t)\alpha''(t)}{\sqrt{1+\alpha'(t)}}=0$$Here we have two possibilities. Either $\alpha'(t)\alpha''(t)=0$ (either $\alpha'(t)=0$ or $\alpha''(t)=0$, both give straight lines, with one a constant function and the other a linear function respectively) or the denominator is infinitely large, which would mean that $\alpha'(t)$ is infinitely large. This means that $\alpha'(t)$ is a vertical line. Q.E.D.
I am skeptical that this proof is true, since in another post by another user (couldn't remember exactly) the proof was using Calculus of Variations and was very long and complicated.