Origin of the major symmetry property of the elasticity tensor

You should start with the strain energy density $\psi$, then define: $$ C_{ijkl} = \frac{\partial^2 \psi}{\partial \epsilon_{ij}\partial \epsilon_{kl}}, $$

and then define $$ \sigma_{ij} = C_{ijkl} \epsilon_{kl} $$

The remainder of my answer will be about explaining why you have to do it that way. Firstly it is physical, there really is energy associated with strain, and if there weren't there would not be any stress. Secondly, it is linear exactly because we are considering the changes in energy due to small strains.

But let's go back to the minor symmetries. We need $C_{ijkl}=C_{jikl}$ because otherwise $\sigma_{ij}\neq\sigma_{ji}$ (and then we get arbitrarily large, hence unphysical, angular velocities for smaller and smaller regions). But the other minor symmetry is not required. If someone handed you a random rank four tensor, let's call it $B_{ijkl}$, and called it an elastic tensor and it didn't have the second minor symmetry, you can define $C_{ijkl}=(B_{ijkl}+B_{ijlk})/2$ $D_{ijkl}=(B_{ijkl}-B_{ijlk})/2$ and then $B=C+D$ but when D is contracted with a symmetric rank two tensor (like the strain tensor) it gives zero. So the part of the rank four tensor without that second minor symmetry simply doesn't contribute, as an actor it does nothing (when you think all the elasticity does is give you stress from strain). So you might as well assume your tensor has both minor symmetries because it acts like it has the second ($B$ and $C$ act the same on symmetric tensors) and it has to have the first.

Did I bring that up to be pedantic? No, I brought it up because the same thing happens if you contract the elasticity tensor with a rank four symmetric combination of the strain tensor. The part of the elasticity tensor without the major symmetry doesn't contribute to the strain energy density. So a random tensor needs the first minor symmetry to be physical. But you might as well assume it has the second minor symmetry since it doesn't affect the stress-strain relationship. And you might as well assume it has the major symmetry because the part that doesn't will not contribute to the strain energy density.

But it is the strain energy density that is physical, and how it changes is what elasticity is. So you aren't really deriving these symmetries as much as saying that only the symmetric ones generate the physical things you want, energy when given strain. And a real derivation should start with strain energy density and strain, and then just define elasticity from that.

Since $\epsilon$ is a symmetric tensor, it has 6 independent components that determine it. Hence use a multi-index $I\in\{(i,j)|1\leq i\leq j\leq 3\}$ to denote them. The strain energy density then becomes (perhaps one has to be careful with "diagonal" terms here in order to get the right coefficients) $$\psi = C_{IJ}\epsilon_I\epsilon_J$$ where summation is intended on repeated multi-indices. It is clear that, since you assume that the order of the second derivatives w.r.t the $\epsilon_I$s doesn't matters, it follows that $$C_{IJ} = C_{JI},$$ for any pair of multi-indices $I$ and $J$. So, in the end, this is the same argument that shows that an Hessian matrix is symmetric if the order of the second derivatives doesn't matter.