$\sum a_i \ln(b_i) \leq \sum a_i \ln(a_i)$ with $\sum a_i = \sum b_i = 1$

Solution 1:

The magic words are "lagrange multipliers". You want to optimize $$f(x_1, \dotsc, x_n)=\sum a_i \log x_i,$$ subject to the constraint $\sum x_i = 1.$

So, you must have $\frac{\partial f}{\partial x_i} = c,$ for some constant $c,$ so $\frac{a_i}{x_i}= c.$ In other words, the $b_i$ at the optimum are proportional to the $a_i,$ but the sums are the same, so $b_i \equiv a_i.$

Solution 2:

By the AM/GM inequality with weights $a_i$, $$ \prod_{i=1}^n \Big(\frac{b_i}{a_i}\Big)^{a_i} \le \sum_{i=1}^n a_i \Big(\frac{b_i}{a_i}\Big) = 1 $$ Rearranging yields $$ \prod_{i=1}^n b_i^{a_i} \le \prod_{i=1}^n a_i^{a_i} $$ and taking logs yields the desired inequality.