Show that $\max(x,by)\ge Ax+By$ for all $x,y$ if and only if $A\in[0,1]$ and $B=(1-A)b$ Let $b,A,B\in\mathbb R$. Can we show that $$\max(x,by)\ge Ax+By\;\;\;\text{for all }x,y\in\mathbb R\tag1$$ if and only if $A\in[0,1]$ and $B=(1-A)b$?
For example, in the "$\Rightarrow$" direction, we may use that $2\max(x,by)=x+by-|x-by|$ to btain that $(1)$ is equivalent to $$(1-2A)x+(b-2B)y\ge|x-by|\;\;\;\text{for all }x,y\in\mathbb R.\tag2$$ Does this imply the claim?
 A: First of all note that we have
$$2\max(x,by)=\max(x,by)+\min(x,by)+|x-by|=x+by+|x-by|.$$
If we take $x=by$ we find
$$2x=2\max(x,by)\geq 2Ax+2By=2(A+\frac{B}{b})x$$
so for $x<0$ we find $A+\frac{B}{b}\geq 1$ and for $x>0$ we find $A+\frac{B}{b}\leq 1$ from which it follows that $B=(1-A)b$. Now suppose $x>by$, if $A>1$ we find
$$\max(x,by)=x\geq Ax+By=Ax+(1-A)by>Ax+(1-A)x=x$$
which is a contradiction, so $A\leq 1$. If $by>x$ and $A<0$ we find
$$\max(x,by)=by\geq Ax+(1-A)by>Aby+(1-A)by=by$$
which is a contradiction, so $A\geq0$.
For the other direction note that $Ax+By=Ax+(1-A)by$ with $A\in[0,1]$. Since a convex combination of two numbers is smaller than their maximum we are done.
A: Only If:
$$y=0 \Longrightarrow \max(x,0) \geqslant Ax \Longrightarrow \left\{ \begin{array}{c} x>0 \Longrightarrow x \geqslant Ax \Longrightarrow 1 \geqslant A \\ x<0 \Longrightarrow 0 \geqslant Ax \Longrightarrow 0 \leqslant A \\ \end{array} \right\} \Longrightarrow A \in [0,1]$$
$$x=by \Longrightarrow by \geqslant Aby+By \Longrightarrow \left\{ \begin{array}{c} y>0 \Longrightarrow b \geqslant Ab+B \\ y<0 \Longrightarrow b \leqslant Ab+B \\ \end{array} \right\} \Longrightarrow B=(1-A)b$$
If:
In fact, when $A$ changes, $[Ax+(1-A)by]$ produces the lines that goes through two points $x$ and $by$. And specially For $A \in [0,1]$, it produces just the line segment between them. In $1$ dimensional spaces that $x,by$ are numbers, it's a weighted mean of them that obviously stays always between them, so we have:
$$\min(x,by) \leqslant Ax+(1-A)by \leqslant \max(x,by)$$
That completes the proof.
