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Prove that

$$f\left( x\right) =\dfrac {1-x^{n}} {\left( 2-x\right) ^{n}-1}, \quad x\in [0,1)$$

is convex for any $n\in \mathbb{N}$.

The second derivative is very unfriendly for a calculus...

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I tried to compute the second derivative and the statement is equivalent to show that some huge polynomial expression (that I believe has lots of cancellations, so is not that huge in the end) has non-negative values in the interval $[0,1)$. I don't know if the polynomial has any zeros in there, but if it doesn't it should be pretty. Have you tried that out? – Patrick Da Silva Aug 19 '12 at 22:57
Try writing it as $f(x)((2-x)^n-1) = 1-x^n$ and use implicit differentiation. – marty cohen Aug 19 '12 at 22:59
The second derivative is not entirely attractive, but not too bad if one resists the urge to expand. – André Nicolas Aug 19 '12 at 23:13
Can somebody show that $x-\frac{1-x^n}{(2-x)^n-1}+\frac{1}{2^n-1}>0$? I think this proves it. – Seyhmus Güngören Aug 19 '12 at 23:19

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