Let $\phi(x) = x\log x$. This function is convex in the set $[0, 1]$.
The entropy $S$ is defined as:
$$S(p) = -\sum \phi(p_i),$$
where $p=[p_1, p_2, \ldots, p_N]$.
Then, using the Jensen's inequality, you get that:
$$ \phi \left(\frac{\sum{p_i}}{N}\right)\leq \frac{\sum{\phi (p_i)}}{N} \Rightarrow \phi \left(\frac{\sum{p_i}}{N}\right)\leq -\frac{S(p)}{N} \Rightarrow S(p) \leq -N \phi \left(\frac{\sum{p_i}}{N}\right).$$
Notice that, whichever is $p$, then by definition $\sum p_i = 1$, and hence:
$$S(p) \leq -N\phi\left(\frac{1}{N}\right) = -N\left(\frac{1}{N}\log\frac{1}{N}\right) = \log N.$$
This means that the entropy is at most equal to $\log N$.
Now, notice that $S(p) = \log(N)$ for $p = \left[\frac{1}{N}, \ldots, \frac{1}{N}\right]$.
Then: $$p_i = \frac{1}{N} \forall i \implies S(p) ~\text{is maximum}.$$
We conclude the proof by observing that the maximum $p_i = \frac{1}{N} ~\forall i$ is unique since $S(p)$ is strictly concave.
The mean in this case is between two terms:
$a_i^2/A^2$ and $b_i^2/B^2$. If you write the AM-GM inequality for these terms for any given $i\in\{1,2,3...,n\}$, you get
$\sqrt{ a_i^2/A^2 \times b_i^2/B^2 }\leq \frac{1}{2}(a_i^2/A^2 + b_i^2/B^2) $.
Summing both sides of this inequality across all $i$ preserves the inequality, and the resulting right hand side simplifies to one, giving the desired crazy sigma equation.
Best Answer
Problem: Assume that we know the following:
$\sum p_i\log \frac{p_i}{q_i} \ge 0$ for $p_i>0 , q_i>0, \forall i$ and $\sum p_i = 1, \sum q_i = 1 $.
Prove that $\prod x_i^{a_i} \le \sum a_i x_i$ for $x_i> 0, a_i > 0, \forall i$ and $\sum a_i = 1$.
Solution: By taking logarithm on both sides, the inequality to be proved is written as $$\sum\nolimits_i a_i\log x_i \le \log \sum\nolimits_i a_i x_i = (\sum\nolimits_i a_i)\log \sum\nolimits_i a_i x_i = \sum\nolimits_i a_i\log \sum\nolimits_j a_j x_j $$ or $$\sum\nolimits_i a_i \Big(\log \sum\nolimits_j a_j x_j - \log x_i\Big)\ge 0$$ or $$\sum\nolimits_i a_i \Big(\log a_i - \log \frac{a_ix_i}{\sum\nolimits_j a_j x_j}\Big)\ge 0.\tag{1}$$ Let $$p_i = a_i, \quad q_i = \frac{a_ix_i}{\sum_j a_j x_j}, \quad i=1, 2, \cdots, n.$$ We know that (1) holds.
We are done.