How do we have the \$E=\frac{1}{2}L_1I_1^2+\frac{1}{2}L_2I_2^2 \pm MI_1I_2\$? Can anyone show me its proof?
Remember the definition of self inductance L is the ratio of the flux linkage generated to the current that generates it

$$
L = \frac{\phi_b}{I}
$$

Some energy is stored in the magnetic field, and as I think you are familiar, this is given by 

$$
E_{single\ inductor} = \frac{1}{2}LI^2
$$

If the two inductors were not coupled the total energy stored would thus be just the sum of two terms like the formula above.

Intuitively, when they are coupled the total energy stored will be less than in the uncoupled situation because some energy from the fields goes into inducing an emf in the other conductor.

Because they are coupled together and their polarity dots are on the opposite sides the currents flowing throw them will lead to an emf that opposes the emfs their own currents generate (someone sanity check this is correct please). 

So the voltage across each inductor will not be the usual 

$$
V= L\frac{dI}{dt}
$$

but instead for one inductor, 

$$
V = L\frac{dI_1}{dt} - M\frac{dI_2}{dt}
$$

and the same thing with the currents swapped for the other

So for one inductor we can find energy by first finding power through VI

$$
\begin{align}
\frac{dE}{dt} &amp;= LI_1\frac{dI_1}{dt} - MI_1\frac{dI_2}{dt} \\
dE &amp;= LI_1 dI_1 - MI_1dI_2 \\
E &amp;= \frac{1}{2}LI_1^2 - M\int I_1dI_2
\end{align}
$$

And for the other inductor 

$$
\begin{align}
\frac{dE}{dt} &amp;= LI_2\frac{dI_2}{dt} - MI_2\frac{dI_1}{dt} \\
dE &amp;= LI_2 dI_2 - MI_2dI_1 \\
E &amp;= \frac{1}{2}LI_2^2 - M \int I_2dI_1
\end{align}
$$

So the total energy is 

$$
E =\frac{1}{2}LI_1^2 + \frac{1}{2}LI_2^2  - M \int I_2dI_1+ I_1dI_2
$$
Since
$$
\frac{d(I_1 I_2)}{dt} =  I_2\frac{d(I_1)}{dt} + I_1 \frac{d(I_2)}{dt}\\
d(I_1 I_2)=  I_2d(I_1) + I_1 d(I_2)\\
\int I_2dI_1+ I_1dI_2 = \int d(I_1 I_2) = I_1 I_2
$$

The total energy can be rewritten as

$$
E =\frac{1}{2}LI_1^2 + \frac{1}{2}LI_2^2  - M I_2I_1
$$

This should help point you in the right direction I hope