Why is an entropy of $\text{log}(n)$ only compatible with the uniform distribution

Question

I have a random variable $X$ and want to show that having an entropy $$ H(X) = - \sum_{i=1}^n p_i \text{log}(p_i) = \text{log}(n)$$ is equivalent to the distribution of $X$ being uniform.

Starting with the distribution is straightforward but I don't see how I can deduce the other implication.

Answer 1

Consider $e^{H(x)} = \prod (\frac{1}{p_i})^{p_i}$

By weighted AM-GM Inequality

$\prod (\frac{1}{p_i})^{p_i} \leq \sum \frac{1}{p_i} \times p_i$ $\implies$ $\prod (\frac{1}{p_i})^{p_i} \leq n$

For the equality to hold: $\frac{1}{p_i} = \frac{1}{p_j}$ for any $i,j$ or equivalently $p_i = p_j$.

And hence $$p_i=\frac{1}{n}$$

Answer 2

Use Jensen's inequality with $f(t) = \log t$. $$H(X) = \sum_{i=1}^n p_i \log \frac{1}{p_i}\leq \log\left(\sum_{i=1}^n p_i \frac{1}{p_i}\right) = \log n$$ $\log t$ is strictly concave, so equality requires $p_i = 1/n$, for all $i$, (in other words, a uniform distribution).