I want to compute $$\max_{\frac{1}{5 \theta }\leq \alpha \leq \frac{1}{2}} \left(\frac{\alpha\log (\alpha)}{1-\alpha} + \log \left( 1 - \alpha\right) + \frac{1}{1-\alpha} \cdot \left( f\left(\frac{1-\alpha}{2\alpha}\right) + 1 + \frac{41}{15\theta} \right)\right)$$ for \begin{align*} f(x) = \begin{cases} - 1.3- \log x + x/2 +1/2 - 1/x&\text{for } x \geq 2\\ -1.3 &\text{otherwise}. \end{cases} \end{align*}

Intuitively, it should be either for $\alpha=1/2$ or for $\alpha = 1/(5\theta)$ (and a plot verifies this). Is there a way how to prove this?

In the end I want to compute the $\theta$ where the whole expression reaches its minimum. Could there be another approach to do this?

Thanks in advance!!

I want to compute $$\max_{\frac{1}{5 \theta }\leq \alpha \leq \frac{1}{2}} \left(\frac{\alpha\log (\alpha)}{1-\alpha} + \log \left( 1 - \alpha\right) + \frac{1}{1-\alpha} \cdot \left( f\left(\frac{1-\alpha}{2\alpha}\right) + 1 + \frac{41}{15\theta} \right)\right)$$ for \begin{align*} f(x) = \begin{cases} - 1.3- \log x + x/2 +1/2 - 1/x&\text{for } x \geq 2\\ -1.3 &\text{otherwise}. \end{cases} \end{align*}

Here, $\log$ is base 2. Intuitively, it should be either for $\alpha=1/2$ or for $\alpha = 1/(5\theta)$ (and a plot verifies this). Is there a way how to prove this?

In the end I want to compute the $\theta \geq 1$ where the whole expression reaches its minimum. Could there be another approach to do this?

Thanks in advance!!