Understanding an inequality with supremum of minumum Let $A,B,C,D$ positive numbers, $g$ be a positive function, $\alpha\in(0,1)$ and the condition $Ag(x)+B\le g(x)^{\alpha}$ holds. Then the following is true:
$$AC+DB\le \min\left[ Cg(x)^{\alpha-1},Dg(x)^{\alpha} \right] 
\le\sup_{t>0}  \min\left[ Ct^{\alpha-1},Dt^{\alpha} \right] 
=C^{\alpha}D^{1-\alpha}.
$$
I am trying to understand these inequalities and equality. But i didn't manage it. It looks very complicated. Can you help me?
 A: Edit: I found a mistake in my solution (step 1), which I will describe there.
Step 1: Proving
$AC+DB\le \min\left[ Cg(x)^{\alpha-1},Dg(x)^{\alpha} \right]$. Equivalently,
$$ AC+DB\le
\begin{cases}
Cg(x)^{\alpha-1}\le Dg(x)^{\alpha},  & D\ge C/g(x) \\
Dg(x)^{\alpha}\le Cg(x)^{\alpha-1}, & D\le C/g(x)
\end{cases}$$
Suppose $AC+DB>Cg(x)^{\alpha-1}$ when $D\ge C/g(x)$. Then $\color{red}{Dg(x)(A-g(x)^{\alpha-1})\ge C(A-g(x)^{\alpha-1})>-DB}$
, which implies that $Ag(x)+B> g(x)^\alpha$, which contradicts our initial assumption that $Ag(x)+B\le g(x)^\alpha$. Thus, the statement is true when $D\ge C/g(x)$. $\color{red}{\text{The issue with these inequalities is that}A-g(x)^{\alpha-1}\text{is necessarily negative.}}$
Step 2: Proving $\min\left[ Cg(x)^{\alpha-1},Dg(x)^{\alpha} \right] 
\le\sup_{t>0}  \min\left[ Ct^{\alpha-1},Dt^{\alpha} \right]$. This is the simplest step, and follows from the definition of $\sup$ and the fact that $g$ is a positive function.
Step 3: Proving $\sup_{t>0}  \min\left[ Ct^{\alpha-1},Dt^{\alpha} \right] 
=C^{\alpha}D^{1-\alpha}$ is a little tricky, and a graph helps. We take advantage of the fact that $\alpha-1<0<\alpha$. $t^\alpha$ will monotonically increase from the origin in an unbounded fashion, and $t^{\alpha-1}$ will monotonically decrease from the $t=0$ asymptote and tend to zero. Thus, these functions will intersect once for $t>0$. $\min\left[ Ct^{\alpha-1},Dt^{\alpha} \right]$ represents $Dt^{\alpha}$ until the point of intersection, and $Ct^{\alpha-1}$ after. Thus, $\sup_{t>0}  \min\left[ Ct^{\alpha-1},Dt^{\alpha} \right]$ represents the value at the point of intersection, which we can now solve for...
A: This is only an observation / clarification that is too long for a comment.A solution has already been provided in the answer of @DavidRaveh.
I will assume that the numbers $A,B,C,D$ are restricted to be strictly positive. Otherwise we could choose $B=0$ and $A$ large to make the inequality
$$
  AC + DB \leq C^\alpha D^{1-\alpha} \tag{1} \label{ineq}
$$
to be violated.
I initially thought that the inequality \eqref{ineq} can not be true, because it's left side depends on $A$ and $B$ whilst the right side does not, so choosing $A$ and $B$ high enough would violate the inequality \eqref{ineq}. However, the existence of a function $g$ gives us an upper bound on the values of $A$ and $B$: Notice that the inequality condition
$$
  A g(x) + B \leq g(x)^\alpha \label{cond} \tag{2}
$$
is satisfied whenever $g(x)\in [f_1(A,B),f_2(A,B)]$, where $f_{1,2}(a,b)$ are the two strictly positive roots of the equation $At+B=t^\alpha$, if they exist. This follows from the fact that $t\mapsto At+B$ is a linear function and $t\mapsto t^\alpha$ is a concave function. The two roots exist iff the linear function $t\mapsto At+B$ is not above the concave function $t\mapsto t^\alpha$ (and then also the function $g$ can exist), which gives us an upper bond on $A$ and $B$ that can be described as
$$
At+B \leq t^\alpha
$$
for some $t>0$.
