<s>[INST] Inverse of this function I have given 
$$ f(x) = \frac{1}{x}\left(\left(\frac{1}{x}\right)^{a} + 1\right)^{-\frac{1}{a}}$$
And am trying to invert it, but despairing.
Perhaps it helps (but I wouldn't know) that $f$ comes from 
$$
g(m,n) = mn (m^a + n^a)^{-1/a}\\
f(x) = g(1/x, 1)\\
$$
 [/INST] Hint:
$f(x) = \frac{1}{x} \bigg( \bigg( \frac{1}{x} \bigg)^a +1 \bigg)^{-\frac{1}{a}}$
$= \bigg( \bigg( \frac{1}{x} \bigg)^a \bigg)^{\frac{1}{a}} \bigg( \bigg( \frac{1}{x} \bigg)^a +1 \bigg)^{-\frac{1}{a}} $
$ = \bigg( \frac{ \big( \frac{1}{x} \big)^a}{\big( \frac{1}{x} \big)^a+1} \bigg)^{\frac{1}{a}}$
$=\bigg( 1- \frac{ 1}{\big( \frac{1}{x} \big)^a+1} \bigg)^{\frac{1}{a}}$
Now you only have one $x$ in the formula. Proceed.
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