Reluctance and inductance are opposites, but both store magnetic energy? Is there a (relatively) simple way to explain this?  Inductance is the creation of stored magnetic energy that resists a change in current direction, but reluctance resists the creation of a magnetic field?  But also stores magnetic energy?
 A: The fact that some entity resists a force (or a field), does not mean that it cannot store energy associated with the action of that force. In fact, this is often (if not always) the case, since a force typically has to perform some work on an entity, presumably, overcoming some resistance, in order to energize it. 
For instance, a spring resists a compression force, but ends up storing energy created by that very force.
Now let's try to apply this rule to magnetic field.
Magnetic field is created by current. 
As you've correctly noted, the inductance resists the increase of the current but we can (loosely) say that the inductor ends up storing magnetic field and, therefore, magnetic energy. This clearly follows the rule outlined above.
At any given current level, the reluctance decreases the magnetic flux and therefore the inductance, which is the ratio of the flux and the current. Therefore, a magnetic circuit with higher reluctance should have lower resistance to the growing current and will end up producing a weaker magnetic field with less magnetic energy. This is, again, consistent with the above rule.  
A: 
Reluctance and inductance are opposites

Inductance $L$ relates a certain magnetic flux linkage $\lambda$ to an electric current $i$ through an $N$ turn coiled conductor:
$$\lambda(i) = N\Phi(i) = Li $$ 
Reluctance $\mathcal{R}$, on the other hand, relates a certain magnetomotive force (mmf) $\mathcal{F}$ to a magnetic flux through a magnetic core.
$$\mathcal{F}(\Phi) = \mathcal{R}\Phi $$
Are reluctance and inductance opposites?  Well, let's see...
If $N$ turns of a conductor are wound around a portion of a magnetic core and the conductor has a current $i$ through, the mmf is just the product of the number of turns and the current:
$$\mathcal{F}(i) = Ni\,[\mathrm{ampere-turns}]$$
The magnetic flux is then
$$\Phi(i) = \frac{\mathcal{F}(i)}{\mathcal{R}} = \frac{N}{\mathcal{R}}i = \frac{L}{N}i$$
Thus,
$$\mathcal{R} = \frac{N^2}{L}$$
Define the permeance $\mathcal{P}$ as
$$\mathcal{P} = \frac{1}{\mathcal{R}}$$
and then
$$\mathcal{P} = \frac{L}{N^2} $$
That is, the permeance is proportional to the inductance.  Now, the energy $W$ stored in the magnetic field of an inductance $L$ due to a current $i$ is given by
$$W = \frac{1}{2}Li^2 = \frac{1}{2}\mathcal{P}N^2i^2 = \frac{1}{2}\mathcal{P}\mathcal{F}^2$$
In summary:
(1) from the electric circuit perspective, the inductance relates the energy stored in the magnetic field to the square of the current.
(2) from the magnetic circuit perspective, the permeance relates the energy stored in the magnetic field to the square of the mmf.
A: I think you are right, energy is stored in inductance as well as reluctance.
Ref:


Energy stored is proportional to overall inductance of magnetic circuit. The overall inductance is primarily governed/limited by the section of magnetic circuit with highest reluctance.
So if you have a long solenoid, you can prove that the reluctance is higher inside the solenoid. ( ext flux density =0, ext reluctance = 0 since flux has infinite area). Hence the energy is stored inside the solenoid.

Now if you have an inductor with a small air gap, the overall reluctance is primarily due to air gap ( since flux density in air gap is almost same as flux density in core). Hence bulk of the energy will be stored in airgap.

PS: In the previous answers, the current was assumed to be constant while comparing the energy storage. We should infact consider flux as being fixed (for a given core size, max permissible flux is fixed) and calculate the energy storage for different reluctances. It can be proven that upon increasing the air-gap, the inductance decreases whereas max energy storage capacity increases- since the current plays a dominating effect over inductance in the equation $\frac{1}{2}L{i}^2$
Ref: Pic courtesy

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*https://electronics.stackexchange.com/questions/461724/energy-in-transformer-air-gap

*https://www.etcourse.com/news-blog/air-gap-in-magnetic-circuits

*https://physics.stackexchange.com/questions/524518/how-to-demonstrate-the-inductance-of-an-inductore
