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Fix an integer $m\geq 1$. Let $f:(0,1)\to \mathbf{R}$ be defined by $f(x) = 16x^{1-m}\exp \left( \frac{8x}{1-x} \right) .$ Can we determine a lower bound for $f$ in terms of $m$?

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    Take the logarithm $g(x)=\ln f(x)$, and write the equation for its minimum, $g'(x)=0$. This yields a quadratic equation in $x$ (after multiplying through by $x(1-x)^2$). You can solve it explicitly and plug the solution back into your original equation to find the exact minimum.2011-12-06

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Differentiating with respect to $x$ yields

f'(x)=f(x)\left(\frac{1-m}x+8\frac{(1-x)+x}{(1-x)^2}\right)=f(x)\left(\frac{1-m}x+\frac{8}{(1-x)^2}\right)\;.

This is zero if the expression in parentheses is zero, that is, for

$8x+(1-m)(1-x)^2=0\;.$

You can solve this quadratic equation for $x$ to find the extrema. Since the function goes to infinity for $x\to0^+$ and $x\to1^-$, exactly one of the two extrema lies in $(0,1)$. I don't think you'll get a particularly nice form for the function value at the minimum, though you can simplify it somewhat by using

$\frac{8x}{1-x}=-(1-m)(1-x)\;.$