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Given $x>0$, $y>0$ and $x + y =1$, how to prove that $\frac{1}{x}\cdot\log_2\left(\frac{1}{y}\right)+\frac{1}{y}\cdot\log_2\left(\frac{1}{x}\right)\ge 4$ ?

2 Answers 2

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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 $f(x,y)\geq{1\over x y}\log_2 2={1\over x y}\geq 4\ .$

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Hint 1: Rewrite this inequality as:

$-x\log_2 x - (1-x)\log_2 (1-x) \geq 4 x (1-x)$

Both sides of the inequality define concave functions on the interval $[0,1]$. Plot them. Can you show that the graph of the second is always lying below the graph of the other?

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    @brian: What I'd do from there is a full sign analysis (first and second derivative) of the difference between the two functions. It's easy to show that there are two inflexion points, and a bit more work to show that there are 3 extrema (2 maxima and 1 minimum). From there, you can find the shape of the curve of the difference and show that it is always positive.2011-04-15