Proof of strong convexity of trace distance I'm trying to follow the Nielsen and Chuang proof (equation 9.49 of Chapter 9, page 408). I reproduce it here for completeness.
With trace distance defined as $D(\rho, \sigma) = \frac{1}{2}tr(|\rho - \sigma|)$, it is known that there exists a projector $P$, such that $D(\rho, \sigma) = tr(P(\rho-\sigma))$. We wish to prove a result for a probabilistic mixture of states, $\sum_i p_i\rho_i$ and $\sum_i q_i\sigma_i$.
There exists a projector $P$ such that
$\begin{align}
D\left(\sum_i p_i\rho, \sum_i q_i\sigma\right) &= \sum_i p_i tr(P\rho_i) - \sum_i q_i tr(P\sigma_i) \\
&=\sum_i p_i tr(P(\rho_i - \sigma_i)) + \sum_i (p_i - q_i) tr(P\sigma_i) \\
&\leq \sum_i p_i D(\rho_i,\sigma_i) + D(p_i,q_i)
\end{align} 
$
Here, $D(p_i,q_i)$ is the classical probability distance given by $D(p_i,q_i) = \frac{1}{2}\sum_i |p_i - q_i|$ for any pair of probability vectors.
I don't understand the very last step, particularly the second term and how $tr(P\sigma_i)$ disappeared. Why is $D(p_i,q_i) = \frac{1}{2}\sum_i |p_i - q_i| \geq \sum_i (p_i - q_i) tr(P\sigma_i)$?
 A: What you are missing is that 
$$
D(X,Y)=\max_P \mathrm{tr}|P(X-Y)|\ ,
$$
where the maximum is over all projectors $P$.
Then, the inequality follows since you are replacing one maximization -- the whole expression is maximized over one P -- by many optimizations over independent $P$'s, which thus can take more optimal values (together with the triangle inequality):
$$
\max_P \mathrm{tr}|P(\sum X_i)| \le \sum \max_P \mathrm{tr}|PX_i|\ ,
$$
with $X_i$ all the terms being summed in the second line.
In addition, it is used that $\mathrm{tr}[P\sigma_i]\le \mathrm{tr}[\sigma_i]=1$.

Let me add that this variational characterization of the trace norm --
$$
\|X\|_1=\mathrm{tr}|X| = \max_P \mathrm{tr}|PX|
$$
over all projectors $P$ -- is extremely useful, so it is good to keep it in mind.  It is a special case of the also very useful min-max-theorem. 
A: To fully answer your question, to get that
$$ \sum_i(p_i - q_i) \leq D(p_i, q_i) = \frac{1}{2}\sum_i|p_i - q_i|$$
(whereas naively you might be tempted to bound $\sum_i(p_i - q_i)$ by $\sum_i|p_i - q_i| = 2\times D(p_i, q_i)$), you need to use Equation 9.4, namely,
$$ D(p_i, q_i) = \max_{S} \sum_{i\in S}(p_i - q_i),$$
where the maximisation is over all subsets $S \subseteq A$ of the full index set. So we have
$$\sum_i (p_i - q_i) = \sum_{i \in A}(p_i - q_i) \leq \max_{S}\sum_{i\in S}(p_i - q_i) = D(p_i,q_i),$$ 
as required to complete the proof.
