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Algebra: proving an inequality
It turns out that the function f can be simplified as f(x) = d\left(\frac{1+x}{2} \parallel \frac{x}{2}\right) - d\left(\frac{x}{2} \parallel \frac{1+x}{2}\right) where d(p \parallel q) = p \log \frac{p}{q} + (1 - p) \log \frac{1-p}{1-q} is the KL divergence between two Bernoulli...- vers
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- Forum: Calculus and Beyond Homework Help
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Algebra: proving an inequality
Certainly if I can show that f'(x) <=0 then the inequality is proved. However, it seems to me that showing f' being negative is even more difficult than showing f being nonnegative directly.- vers
- Post #3
- Forum: Calculus and Beyond Homework Help
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Algebra: proving an inequality
Homework Statement Let f(x) = (1/2+x) log [(1+x)/x] + (3/2-x) log [(1-x)/(2-x)] where log is natural logarithm and 0 < x <= 1/2. Show that f(x) >= 0 for all x.Homework Equations The only inequality that I can think of is the log sum inequality: http://en.wikipedia.org/wiki/Log_sum_inequalityThe...- vers
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- Algebra Inequality
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- Forum: Calculus and Beyond Homework Help