Is there positive such that the inequality is true? Let $A\in [0,1]$ and $B\in (0,1)$, then there is small $c&gt;0$ such that $A-cB\geq 0$.
Is the statement true ?
My way if A is not zero that is clear, cause I can choose $c=\frac{A}{B}$. Moreover, $\frac{c}{2}$. But in case $A=0$ ?
From what I see, the relation $A-cB\geq 0$ is only true for $c&gt;0$ when $A\in ( 0,1] ,\ B\in [ 0,1]$
Here it is
$ \begin{array}{l}
A-cB\geqslant 0,\\
\\
cB\leqslant A\Longrightarrow c\leqslant \frac{A}{B}\\
\\
0&lt; c\leqslant \frac{A}{B} ,\ \frac{A}{B} \ cannot\ be\ zero\Longrightarrow A\neq 0,\ B\neq \infty ,-\infty \\
\Longrightarrow A\in ( 0,a] ,\ B\in [ b,c] ,\ where\ a,c,b\ are\ both\ finite\ numbers\ with \ a\geq0
\end{array}$
The only way that $0-cB \geq 0$ is that either $c $ or $B $ is equal to zero (because $cB\overset{!}{=}0$).
So no, this is not possible.