TITLE: how to prove this inequality?
QUESTION [5 upvotes]: Given $x>0$, $y>0$ and $x + y =1$, how to prove that $\frac{1}{x}\cdot\log_2\left(\frac{1}{y}\right)+\frac{1}{y}\cdot\log_2\left(\frac{1}{x}\right)\ge 4$ ?

REPLY [7 votes]: The function $t\mapsto\log_2{1\over t}$ is convex. Apply Jensen's inequality to
$$f(x,y):={1\over x}\log_2{1\over y}+{1\over y}\log_2{1\over x}={1\over x y}\Bigl(y\ \log_2{1\over y}+x\ \log_2{1\over x}\Bigr)$$
and obtain
$$f(x,y)\geq{1\over x y}\log_2 2={1\over x y}\geq 4\ .$$