I'm reading a monograph that considers the following problem:

$$\inf_{z(t) \in C^1} \int_0^1 c\bigg(\frac{dz(t)}{dt}\bigg) dt\\ z(0) = x, z(1) = y$$

Here $c$ is a convex function, $z(t)$ are paths with initial and final points given. They claim the infimum is $c(y-x)$ and this follows from Jensen's inequality. I can see part of it:

$$\int_0^1 c\bigg(\frac{dz(t)}{dt}\bigg) dt \geq c\bigg(\int_0^1 \frac{dz(t)}{dt} dt\bigg)=c(z(1)-z(0))=c(y-x)$$

They claim though that:

$$\inf_{z(t) \in C^1} \int_0^1 c\bigg(\frac{dz(t)}{dt}\bigg) dt=c(y-x)$$

How do we know that the inf of LHS is the RHS -- can't the inf end up greater than the RHS?

Edit: As the commenter points out, a constant speed trajectory achieves the inf. I should ask: without knowing this answer ahead of time, I'm having trouble seeing how I could obtain this from this variational problem.

(Specifically, I am looking at this book -- page 145, Prop 5.2)