Find $\min_{x \geq \beta} \beta^{\frac{x-\beta}{\beta-1}} x^{-\beta}$

Question

I am looking for a solution to the following problem, $$ \min_{x \geq \beta} \beta^{\frac{x-\beta}{\beta-1}} x^{-\beta}, \text{ where }\beta > 1. $$ For example, if $\beta=2$, then the problem becomes minimizing $\frac{2^x}{x^2}$, which gives $x^*(2) = \frac{2}{\log 2}$. Numerical calculations seem to indicate that the there is always a unique minimizer $x^*(\beta)$. It seems always to be interior and increasing with $\beta$. However, it would be very useful to find an explicit expression or at least a characterization for $x^*(\beta)$.

Answer 1

$\beta > 1 $ so $\forall x \geq \beta, \ f(x)>0$ and $ \min \limits_{x \geq \beta} f(x) = \min \limits_{x \geq \beta} \log(f(x))$

$g(x)=log(f(x)) = \frac{x-\beta}{\beta -1} \log(\beta) - \beta \log(x) $

$g'(x)=\frac{\log(\beta)}{\beta -1}-\frac \beta x $

$g'(x^*)=0 \iff x^*=\frac{\beta (\beta-1)}{\log(\beta)}$

$g''(x) = \frac \beta{x^2} >0$ so $x^*$ is the global minimum

Answer 2

Putting $f(x) = \beta^{\frac{x - \beta}{\beta - 1}} x^{-\beta}$, we can find $\min \limits_{x \geq \beta} f(x)$ by differentiating $f(x)$. Using the product rule, we find $$ f'(x) = \beta^{\frac{x - \beta}{\beta - 1}} \frac{\ln \beta}{\beta - 1} x^{-\beta} - \beta x^{- \beta - 1} \beta^{\frac{x - \beta}{\beta - 1}} = x^{-\beta - 1} \beta^{\frac{x - \beta}{\beta - 1}} \cdot \left(x \frac{\ln \beta}{\beta - 1} - \beta \right). $$ As we assume that $x \geq \beta > 1$, the first factor is always strictly positive. Hence, solving $f'(x) = 0$, we find that $$ x^* = \frac{\beta (\beta - 1)}{\ln \beta}. $$ Showing that it is in fact a minimum, comes down to showing that $f'(x)$ is negative for $x < x^*$ and positive for $x > x^*$, which follows from the fact that the sign of $f'(x)$ only depends on the factor between brackets, which is a linear function with positive slope. Also, since $\beta > \ln \beta + 1$ for $\beta > 1$, we find that $x^* > \beta$. Hence, $f$ attains a minimum for $x^*(\beta) = \frac{\beta(\beta - 1)}{\ln \beta}$.