sum of log versus log of sum

Question

I am struggling to show the following:

Let $p \in (0,1), q_i \in (0,1)$ for all i = 1, 2, ..., k. Show that

$$ p\log \left(\frac{p}{\frac{1}{k}\sum_{i=1}^kq_i}\right) \leq \frac{1}{k}\sum_{i=1}^kp\log\left(\frac{p}{q_i}\right). $$

Can anyone help me please?

Answer 1

As you have already mentioned above, it is straightforward to rewrite this inequality as $$ \log\left(\frac 1k\sum_{i=1}^kq_i\right) \ge \frac 1k \sum_{i=1}^k\log q_i = \log \sqrt[k]{\prod_{i=1}^kq_i}. $$ Due to the monotonicity of the logarithm function, this is equivalent to $$ \frac 1k\sum_{i=1}^kq_i \ge \sqrt[k]{\prod_{i=1}^kq_i}.$$ This inequality is well-known: AM-GM. The wikipedia article also contains several methods to prove this inequality.

Answer 2

LHS: $$ p\ln \left(\frac{p}{\frac{1}{k}\sum_{i=1}^kq_i}\right) = p \ln p - p\ln\left(\sum_{i=1}^k \frac{q_i}{k}\right) $$

RHS: $$ \frac{1}{k}\sum_{i=1}^kp\ln\left(\frac{p}{q_i}\right) = \frac{p}{k}\sum_{i=1}^k (\ln p - \ln q_i) = p\ln p - p \sum_{i=1}^k \frac{\ln q_i}{k} $$

and you can apply Jensen's inequality since $\ln()$ is convex.