If $p_1,p_2>0$ and $p_1+p_2=1$ then prove that $$p_1\log(\frac{1}{p_1})+p_2\log(\frac{1}{p_2})\leq \log2$$
I`ve tried to define $p_1$ as $1-p_2$, and I got $$(\frac{(1-p_2)}{p_2})^{p_2}*\frac{1}{1-p_2}\leq2$$
But I`m not sure what to do next, and I don't know if that's the right way.
Could you give me any hints? 🙂
Best Answer
Using the logarithm rules it suffices to show $$ \log(p_1^{-p_1}p_2^{-p_2})\leq \log(2), $$ and since $\log$ is increasing on its domain of definition it suffices to show $$ p_1^{-p_1}p_2^{-p_2}\leq 2. $$ By Young's product inequality (https://en.wikipedia.org/wiki/Young%27s_inequality_for_products) we now have $$ p_1^{-p_1}p_2^{-p_2}\leq p_1p_1^{-1}+p_2p_2^{-1}\leq 2. $$