There was a question about comparing two entropies and showing that one of them is greater than or equal to other. After getting rid of the same terms for both expressions, I am left with the following expressions:
$$A = p_i\log_2\bigg(\frac{1}{p_i}\bigg)+p_j\log_2\bigg(\frac{1}{p_j}\bigg)$$
and
$$B = (p_i+p_j)\log_2\bigg(\frac{2}{p_i+p_j}\bigg)$$
where $p_i$ and $p_j$ are probabilities ($0 \le p_i,p_j \le 1$). Here, I need to show that $B \ge A$.
First, I thought about Jensen's Inequality because expression $A$ can be expressed as an expectation of a logarithm function $\mathbb{E}[f(X)]$. But then I could not express the expression $B$ as $f(\mathbb{E}[X])$ where I took $f$ as $\log_2\big(\frac{1}{x}\big)$. In the end, I could not use Jensen's Inequality.
Second, I thought that result may follow just by the definition of convexity. Therefore, I manipulated $B$ a little to have:
$$B = (p_i+p_j)\bigg[1-\log_2\bigg(\frac{1}{p_i+p_j}\bigg)\bigg] = (p_i+p_j)-(p_i+p_j)\log_2\bigg(\frac{1}{p_i+p_j}\bigg)$$
Then, I need to show that
$$p_i+p_j \ge p_i\log_2\bigg(\frac{1}{p_i}\bigg)+p_j\log_2\bigg(\frac{1}{p_j}\bigg) + (p_i+p_j)\log_2\bigg(\frac{1}{p_i+p_j}\bigg)$$
Here, I am stuck. Any help or suggestion is appreciated. Thank you in advance.
$B \ge A$ is equivalent to $$ \frac{p_i+p_j}{2} \log_2 \left(\frac{p_i+p_j}{2}\right) \le \frac 12 \left( p_i \log_2 p_i + p_j \log_2 p_j\right) $$ and that is true because the function $f(x) = x \log_2(x)$ is convex for $x > 0$: $f''(x) = \frac {1}{x \log 2} > 0$.