Assume S is a regular surface with a Riemannian metric $g$.
The energy of a curve $c: [0,a] \to S$ is defined as :
$$ E(c) = \frac{1}{2} \int_0^a g_{c(t)}(\dot c(t),\dot c(t))\mathsf{dt} $$
This is quite similar to the definition of the length in terms of the defining functions i.e.:
$$L(c) = \int_0^a \sqrt{g_{c(t)} (\dot c(t), \dot c(t) )}\mathsf{dt}$$
The Lemma 2.3 in Chapter 9 of Riemannian Geometry by Do Carmo is:
About the formula $$aE(\gamma)=(L(\gamma))^2\leq (L(c))^2\leq aE(c),$$ the author has proved the inequality: $$(L(c))^2\leq aE(c)$$ using the Schwarz inequality.
The question is: why $aE(\gamma)=(L(\gamma))^2$?
For any smooth curve $c : [0,a] \to M$, choosing $f \equiv 1$ and $g=|c'(t)|$ the Cauchy-Schwarz inequality $$ \Big( \int_0^a fg \,\, \text{d}t\Big)^2 \leq \int_0^a f^2 \, \text{d}t \int_0^a g^2 \, \text{d}t, $$ become $L(c)^2 \leq aE(c)$. But Cauchy-Schwarz inequality become an equality if and only if $g$ is constant, which means that $L(c)^2 = aE(c)$ if and only if $|c'(t)| = \text{const}$.
Apply this inequality to the curve $\gamma$ in the lemma, we certainly have $L(\gamma)^2 \leq aE(\gamma) $. Since by hypotheses, $\gamma$ is a geodesic ( i.e. $D\gamma'/dt = 0$ ), a simple computation $$ \frac{d}{dt} \langle \gamma',\gamma' \rangle = \Big\langle \frac{D\gamma'}{dt}, \gamma'\Big\rangle + \Big\langle \gamma', \frac{D\gamma'}{dt} \Big\rangle = 0, $$ implies the speed $|\gamma'(t)|$ is constant. Therefore $L(\gamma)^2 = aE(\gamma)$.