is there a function $gamma(x)$ where when $a$ & $b$ and $a+1$ & $b+1$ are co-prime, $gamma(frac{a}{b})gamma(frac{a+1}{b+1})$

Question

is there a function $gamma(x)$ where when $a$ & $b$ and $a+1$ & $b+1$ are co-prime, $gamma(frac{a}{b})>gamma(frac{a+1}{b+1})$
when you start with $gamma(frac{1}{2})$ you get an inequality for all n
$$gamma(frac{1}{2})>gamma(frac{2}{3})>gamma(frac{3}{4})>...>gamma(frac{n}{n+1})>gamma(frac{n+1}{n+2})>...$$
but if you start with $frac{1}{3}$ you find no new information because $frac{1+1}{3+1}$ isn't fully simplified so $3+1$ and $1+1$ isn't coprime so we don't know if $gamma(frac{1}{3})>$or$=$ or$<gamma(frac{1}{2})$
is there a function that follows this rule for all fractions $frac{a}{b}$ and that's differentiable everywhere
And if there is a function $gamma(x)$ then is it made up from elementary functions?

Martin R · Answer

Assuming that $a, b$ are positive integers: $$ frac{a}{b} < 1 implies frac{a}{b} < frac{a+1}{b+1} < 1 \ frac{a}{b} > 1 implies frac{a}{b} > frac{a+1}{b+1} > 1 \ $$ (compare also How to prove that adding $n$ to the numerator and denominator will move the resultant fraction close to $1$?). Therefore any function $gamma$ which is strictly decreasing on $(0, 1)$ and strictly increasing on $(1, infty)$ has the property that $$ gamma(frac{a}{b})>gamma(frac{a+1}{b+1}) $$ for all $a ne b$, and in particular for co-prime integers. There are many such functions, a simple example is $gamma(x) = x + frac 1x$.

is there a function $gamma(x)$ where when $a$ & $b$ and $a+1$ & $b+1$ are co-prime, $gamma(frac{a}{b})>gamma(frac{a+1}{b+1})$

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