maximum value of $\sum_i a_i \log b_i $ when $\sum a_i = \sum b_i = 1$

Question

$J=\sum_i a_i \log b_i $

and $\sum_i a_i = 1, \sum_i b_i = 1 $

when $a_i = b_i $, $J$ has a maximum value.(Is this true?)

How to prove this?

Answer 1

Consider two families $a=(a_1,\dots,a_N)$ and $b=(b_1,\dots,b_N)$ such that $\sum_{i=1}^Na_i=\sum_{i=1}^Nb_i=1$.

Then, we claim that $\sum_{i=1}^Na_i\ln(b_i)\leqslant\sum_{i=1}^Na_i\ln(a_i)$ (*) with equality if and only if $a_i=b_i$ for any $i=1,\dots,N$.

For, we have $\sum_{i=1}^Na_i\ln(b_i)-\sum_{i=1}^Na_i\ln(a_i)=\sum_{i=1}^Na_i\ln(\frac{b_i}{a_i})$ but, by a study of the function $\ln(x)-x+1$, we know that $\ln(x)\leqslant x-1$ for any $x>0$, and the equality occurs if and only if $x=1$.

So $\sum_{i=1}^Na_i\ln(b_i)-\sum_{i=1}^Na_i\ln(a_i)\leqslant \sum_{i=1}^na_i(\frac{b_i}{a_i}-1)=\sum_{i=1}^Nb_i-\sum_{i=1}^Na_i=0$

Thus, if $J$ is maximum, then $a_i=b_i$ for every $i$.