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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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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