Understanding the proof of the log sum inequality. 
In the proof above I find that the 2.101 is too sketchy to understand. And why can the $b_i$ relate to $a_i$ by $a_i=\frac{a_i}{\sum_{j=1}^nb_j}$ ? And $t_i=\frac{a_i}{b_i}$? 
Please point me in the right direction. Thanks very much. 
 A: When questions like this arise, I find it helpful to try to re-write the inequality with equality, plus some remaining terms. That is, if you want to prove that
$X \geq Y$, try to find an expression $X = Y + Z$ for some $Z$ and prove that $Z \geq 0$. That way, the steps in the proof become more apparent and self-motivating.
To see how this would work for the log-sum inequality, let
$$
A = \sum_{i=1}^{n} a_{i}, B = \sum_{i=1}^{n}b_{i}, \tilde{a}_{i} = \frac{a_{i}}{A}, \tilde{b}_{i} = b_{i}/B.
$$
Then,
\begin{align*}
    \sum_{i=1}^{n} a_{i}\log\frac{a_{i}}{b_{i}}
    &= A \cdot \left( \sum_{i=1}^{n} \tilde{a}_{i} \log \left[
      \frac{\tilde{a}_{i}}{\tilde{b}_{i}} \cdot \frac{A}{B} \right]
    \right) \\
    &= A \cdot \left( \sum_{i=1}^{n} \tilde{a}_{i} \log
            \frac{\tilde{a}_{i}}{\tilde{b}_{i}}
            + \sum_{i=1}^{n} \tilde{a}_{i} \log \frac{A}{B}
    \right) \\
    &= A \cdot \left(
      KL(\tilde{a} || \tilde{b}) + \log \frac{A}{B}
    \right) \\
    &= A \cdot KL(\tilde{a} || \tilde{b}) + A \log \frac{A}{B},
  \end{align*}
where $KL(\tilde{a} || \tilde{b})$ denotes the KL-divergence between the
probability distributions $(\tilde{a}_{i})_{i=1,\dots,n}$ and $(\tilde{b}_{i})_{i=1,\dots,n}$.
By convexity of $x \mapsto x \log x$, KL-divergence is always non-negative, and equal to zero if and only if $\tilde{a}_{i} = \tilde{b}_{i}$ for all $i = 1, \dots, n$. This completes the proof.
A: Since the given choices of $\alpha_i,\,t_i$ imply $\alpha_i\ge0$, $\sum_i\alpha_i=1$ and$$\sum_i\alpha_it_i=\sum_i\frac{b_i}{\sum_jb_j}\frac{a_i}{b_i}=\sum_i\frac{a_i}{\sum_jb_j},$$the right-hand side of (2.100) becomes that of (2.101). Similarly, the left-hand side is$$\sum_i\frac{b_i}{\sum_jb_j}\frac{a_i}{b_i}\log\frac{a_i}{b_i}=\sum_i\frac{a_i}{\sum_jb_j}\log\frac{a_i}{b_i},$$as claimed.
