Proof of "Entropy of preparation" in Von Neumann entropy

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To prove the "entropy of preparation" in Von Neumann entropy, start with a pure state drawn from the ensemble {|φx〉, px}, leading to the density matrix ρ = ∑px|φx〉<φx|. The relationship H(X)≥S(ρ) is established, where H is the Shannon entropy of the probabilities {px} and S is the Von Neumann entropy. By expressing |φx〉 in terms of the eigenstates of ρ, it can be shown that S(ρ) is less than or equal to H(X) plus the sum of the weighted entropies of the individual states. This demonstrates that the entropy of a mixture of pure states is greater than the mixture of their entropies, confirming the desired result. The proof can be found detailed on page 519 of Nielsen and Chuang's textbook.
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How should I prove this?

From John Preskill's quantum computation & quantum information lecture notes(chapter 5)

If a pure state is drawn randomly from the ensemble{|φx〉,px}, so that the density matrix is ρ = ∑pxx〉<φx|
Then, H(X)≥S(ρ)
where H stands for Shannon entropy of probability {px} and S stands for Von Neumann entropy.
 
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If you can find it, on page 519 in Nielsen and Chuang's (Quantum Computation and quantum information, 10th anniversary edition), they show how to do it.

In short, you express |\phi_{x}\rangle in terms of the eigenstates of the density matrix \rho. Since the entropy of a mixture of states is larger than the corresponding mixture of entropies, you can eventually find:
S(\rho)\leq H(X)+\sum_{x}p_{x}S(\rho_{x})
Since the states in the ensemble are pure, this would give you your result.
 
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Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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