For a one qubit system, take a basis. Call this the mixture basis. Consider only basis states and classical mixtures of these basis states.
Definition of Shannon Entropy used here: Defined with respect to the measurement basis, on the probabilities of various outcomes. For eg: $\frac{1}{2}|0\rangle \langle0| + \frac{1}{2}|+\rangle \langle+|$, when measured in the $|0\rangle, |1\rangle$ basis has Shannon Entropy $-\frac{3}{4}\log(\frac{3}{4}) - \frac{1}{4}\log(\frac{1}{4})$ because there is $\frac{3}{4}$ chance of measuring $|0\rangle$ and $\frac{1}{4}$ chance of measuring $|1\rangle$.
I'm trying to prove that the least value of Shannon Entropy will occur when the measurement basis is equal to the mixture basis.
(This is for me to get an intuition of Von Neumann entropy. If I prove the above, then I can think of Von Neumann entropy as the least Shannon entropy I could get after measuring across any basis.)
Let the mixture basis be $\frac{1}{2}(I + n.\sigma)$ and $\frac{1}{2}(I - n.\sigma)$
Let the measurement basis be $\frac{1}{2}(I + m.\sigma)$ and $\frac{1}{2}(I - m.\sigma)$
Let the qubit be $$p\frac{1}{2}(I + n.\sigma) + (1-p)\frac{1}{2}(I - n.\sigma)$$
Probability of the qubit showing up as $\frac{1}{2}(I + m.\sigma)$ when measured is: $$p(\frac{1}{2}(1 + m.n)) + (1-p)(\frac{1}{2}(1-m.n))$$
Let the above value be $p^{'}$
Then the Shannon Entropy will be $p^{'}log(p^{'}) + (1-p^{'})log(1-p^{'})$
And to minimize the entropy, I need to minimise or maximise $p^{'}$
I'm not sure how to do that though, and whether what I'm trying to do so far makes sense. I'll be grateful for any help on continuing the proof/ insight on the intuition I'm trying to build.