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

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2 Answers 2

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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Hint 2: raise both sides of the expression Raskolnikov wrote to the exponent with base 2. –  Willie Wong Apr 14 '11 at 16:38
    
Thanks for comment. Can you provide some details, 'cause I didn't get it. –  user9587 Apr 15 '11 at 2:30
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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. –  Raskolnikov Apr 15 '11 at 9:21

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