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!!



Computing minimum / maximum of strange two variable funcion
Let $a:=\alpha$ and $u:=\frac1{5\theta}$. The condition $\theta\ge1$ (now added in the question) means that $0<u\le1/5$, which will be assumed henceforth. 
We need to compute 
\begin{equation}
 \inf\_{u\in(0,1/2)}\sup\_{a\in[u,1/2]}F(u,a), 
\end{equation}
where
\begin{equation\*}
 F(u,a):=\begin{cases}
 F\_1(u,a)&\text{ if }1/5\le a\le1/2,\\
 F\_2(u,a)&\text{ if }u\le a\le1/5,
 \end{cases} 
\end{equation\*}
\begin{equation}
 F\_1(u,a):=\frac{30 (a-1) \ln (1-a)-30 a \ln a+(9-410 u) \ln2}{30 (a-1) \ln2},
\end{equation}
\begin{equation}
 F\_2(u,a):=\frac{H(u,a)}{60 (1-a)^2 a \ln2}, 
\end{equation}
\begin{multline}
 H(u,a):=-60 a^3 \ln a-820 a^2 u \ln2+60 a^2 \ln a-117 a^2 \ln2+820 a u \ln2 \\ 
 +60
 (a-1) a \ln \left(\frac{1}{2} \left(\frac{1}{a}-1\right)\right)+60 (a-1)^2 a \ln
 (1-a)-18 a \ln2+15 \ln2. 
\end{multline}
So, the infsup in question is
\begin{equation}
 \inf\_{0<u\le1/5}(M\_1(u)\vee M\_2(u)), 
\end{equation}
\begin{equation}
 M\_1(u):=\sup\_{1/5\le a\le1/2}F\_1(u,a),\quad
 M\_2(u):=\sup\_{a\in[u,1/5]}F\_2(u,a). 
\end{equation} 


For $F\_1(a):=F\_1(u,a)$, let $DF\_1(a):=F\_1'(a)(1-a)^2$. Then $DF\_1(a)=-3/10 + 41 u/3 + \ln a/\ln2$ is increasing in $a$. So, $DF\_1(a)$ (and hence $F\_1'(a)$) can change the sign only from $-$ to $+$. So, 
\begin{align}
 M\_1(u)&=\sup\_{1/5\le a\le1/2}F\_1(u,a)=F\_1(u,1/5)\vee F\_1(u,1/2). 
\end{align}


For $F\_2(a):=F\_2(u,a)$, let $DF\_2(a):=F\_2'(a)(1-a)^2$. Then $DF\_2'(a) 2 \ln2\,(1-a)^2 a^3=2 a^4 + \ln2 - a (2 + \ln8) + a^2 (8 + \ln8) - 
 a^3 (8 + \ln512)>0$ for $a\in[0,1/5]$. So, $DF\_2(a)$ is increasing in $a\in[0,1/5]$, and so, $DF\_2(a)$ (and hence $F\_2'(a)$) can change the sign only from $-$ to $+$. So, 
\begin{equation}
 M\_2(u)=\sup\_{a\in[u,1/5]}F\_2(u,a)=
 F\_2(u,u)\vee F\_2(u,1/5)\quad\text{if }0<u\le1/5. 
\end{equation}


Therefore and because $F\_1(u,1/5)=F\_2(u,1/5)$, the infsup in question is
\begin{equation}
\inf\_{0<u\le1/5}[F\_1(u,1/5)\vee F\_1(u,1/2)\vee F\_2(u,u)]. 
\end{equation}


For brevity, let 
\begin{equation}
 g(u):=F\_2(u,u),\quad d\_a(u):=F\_2(u,u)-F\_1(u,a)=g(u)-F\_1(u,a). 
\end{equation}
Let $g\_1(u):=g'(u)(1 - u)^2$. Then $g\_1'(u)$ is a simple rational function of $u$, which is $>0$ (everywhere here $0<u\le1/5$). So, $g\_1(u)$ is increasing in $u$. So, $g\_1(u)$ (and hence $g'(u)$) can change the sign only from $-$ to $+$. Now we find 
\begin{equation}
 \inf\_{0<u\le1/5}F\_2(u,u)=\min\_{0<u\le1/5}g(u)=1.0616\ldots, 
\end{equation}
attained at $u=0.099677\ldots$. 


Because (i) $d\_a(u)=g(u)-F\_1(u,a)$, (ii) $F\_1(u,a)$ is affine in $u$, and (iii) $g'(u)$ can change the sign only from $-$ to $+$, we see that $d\_a'(u)$ can change the sign only from $-$ to $+$. Also, $d\_{1/5}'(14/100)=-6.7650\ldots<0$. So, $d\_{1/5}$ is decreasing on $[0,14/100]$, with $d\_{1/5}(14/100)=0.1884\ldots>0$. So, $d\_{1/5}>0$ on $[0,14/100]$, that is, $F\_2(u,u)>F\_1(u,1/5)$ if $0<u\le14/100$. Similarly, using that $d\_{1/2}'(14/100)=-17.015\ldots<0$ and $d\_{1/2}(14/100)=0.07602\ldots>0$, we verify that $F\_2(u,u)>F\_1(u,1/2)$ if $0<u\le14/100$. 


So, 
\begin{equation}
\inf\_{0<u\le14/100}[F\_1(u,1/5)\vee F\_1(u,1/2)\vee F\_2(u,u)]= \min\_{0<u\le14/100}F\_2(u,u)= 1.0616\ldots, 
\end{equation}
attained at $u=0.099677\ldots$. 
On the other hand, because $F\_1(u,1/2)$ is increasing in $u$, we have 
\begin{multline}
\inf\_{14/100<u\le1/5}[F\_1(u,1/5)\vee F\_1(u,1/2)\vee F\_2(u,u)]\ge\inf\_{14/100<u\le1/5}F\_1(u,1/2) \\ 
= 
F\_1(14/100,1/2) =1.2266\ldots \\ 
>1.0616\ldots
=\inf\_{0<u\le14/100}[F\_1(u,1/5)\vee F\_1(u,1/2)\vee F\_2(u,u)]. 
\end{multline}



> 
> Thus, with $M(u):=F\_1(u,1/5)\vee F\_1(u,1/2)\vee F\_2(u,u)$, the infsup in question is
>  \begin{multline}
> \inf\_{0<u\le1/5}M(u) 
> =\inf\_{0<u\le14/100}M(u) 
> \bigwedge
> \inf\_{14/100<u\le1/5}M(u)\\
> =\inf\_{0<u\le14/100}M(u)= 1.0616\ldots, 
> \end{multline}
>  attained at $a=u=0.099677\ldots$. 
> 
> 
> 


